JEE Main 22 January 2025 Shift 1 (Previous Year Paper)

Concept Used:

The number of equivalence relations on a set is equivalent to the number of ways the set can be partitioned. For a set containing n elements, this number is represented by the Bell number Bn.

Calculation:

⇒ Partitions of A = {1, 2, 3} are:

⇒ 1. {{1}, {2}, {3}}

⇒ 2. {{1, 2}, {3}}

⇒ 3. {{1, 3}, {2}}

⇒ 4. {{2, 3}, {1}}

⇒ 5. {{1, 2, 3}}

⇒ Number of partitions = 5

⇒ Bell number B3 = 5

⇒ Number of equivalence relations = 5

The empty relation is not included

∴ The maximum number of equivalence relations on A = {1, 2, 3} is 5

Hence option 3 is correct

Calculation 

f(x + y) = f(x).f(y) 

⇒ f(x) = e^λx f′(0) = 4a 

⇒ f′(x) = λe^λx ⇒ λ = 4a

So, f(x) = e^4ax

f′′(x) – 3af′(x) – f (x) = 0 

⇒ λ² – 3aλ – 1 = 0 

⇒ 16a² – 12a² – 1 = 0 ⇒ 4a² = 1 ⇒ $\mathrm{a}=\frac{1}{2}$

F(x) = e^2x

Area = $\int_{0}^{2} \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{e}^{2}-1$

Hence option 1 is correct

Calculation 

Let ‘G’ be the centroid of Δ formed by (1, 3) (3, 1) & (2, 4)

$\mathrm{G} \cong\left(2, \frac{8}{3}\right)$

Image of G w.r.t. x + 2y – 2 = 0

$\frac{α-2}{1}=\frac{β-\frac{8}{3}}{2}=-2 \frac{\left(2+\frac{16}{3}-2\right)}{1+4}$

= $\frac{-2}{5}\left(\frac{16}{3}\right)$

⇒ $α=\frac{-32}{15}+2=\frac{-2}{15}, β=\frac{-32 \times 2}{15}+\frac{8}{3}=\frac{-24}{15}$

15(α – β) = – 2 + 24 = 22

Hence option 4 is correct

Calculation

Given:

$Z_{1}=e^{-i\pi/4}, Z_{2}=1, Z_{3}=e^{i\pi/4}$

\( |z_{1}\overline{z}{2}+z{2}\overline{z}{3}+z{3}\overline{z}_{1}|^{2} = |e^{-i\frac{\pi}{4}}\times1+1\times e^{-i\frac{\pi}{4}}+e^{i\frac{\pi}{4}}\times e^{i\frac{\pi}{4}}|^{2} \)

$= |e^{-i\frac{\pi}{4}}+e^{-i\frac{\pi}{4}}+e^{i\frac{\pi}{4}}|^{2}$

$= |2e^{-i\frac{\pi}{4}}+1|^{2}$

$= |\sqrt{2}-\sqrt{2}i+i|^{2}$

$= (\sqrt{2})^{2}+(1-\sqrt{2})^{2} = 2+1+2-2\sqrt{2} = 5-2\sqrt{2}$

$\alpha=5, \beta=-2$

$\Rightarrow \alpha^{2}+\beta^{2}=29$

Hence option (4) 29 is correct

Calculation 

16(sec^–1 x)² + (cosec^–1 x)²

\operatorname{Sec}^{-1} \mathrm{x}=\mathrm{a} \in[0, π]-\left{\frac{π}{2}\right}

$\operatorname{cosec}^{-1} \mathrm{x}=\frac{π}{2}-\mathrm{a}$

= $16\left[\mathrm{a}^{2}+\left(\frac{π}{2}-\mathrm{a}\right)^{2}\right]=16\left[2 \mathrm{a}^{2}-π \mathrm{a}+\frac{π^{2}}{4}\right]$

$\max ]_{a=π}=16\left[2 π^{2}-π^{2}+π \frac{2}{4}\right]=20 π^{2}$

$\min ]_{a=\frac{π}{4}}=16\left[\frac{2 \times π^{2}}{16}-\frac{π^{2}}{4}+\frac{π^{2}}{4}\right]=2 π^{2}$

Sum = 22π² 

Hence option 4 is correct

Calculation 

HHH → 0

HHT → 0

HTH → 1

HTT → 0

THH → 1

THT → 1

TTH → 1

TTT → 0

Probability distribution

\(\begin{array}{c|c|c} \mathrm{x}{\mathrm{i}} & 0 & 1 \ \hline \mathrm{P}\left(\mathrm{x}{\mathrm{i}}\right) & 1 / 2 & 1 / 2 \end{array}\)

\(\mu=\sum \mathrm{x}{\mathrm{i}} \mathrm{p}{\mathrm{i}}=\frac{1}{2}\)

$\sigma^{2}=\sum x_{i}^{2} p_{i}-\mu^{2}$

= $\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$

$64\left(\mu+\sigma^{2}\right)=64\left(\frac{1}{2}+\frac{1}{4}\right)=48$

Hence option 2 is correct

Calculation

a₁ .a₅ = 28 ⇒ a.ar⁴ = 28 ⇒ a² r⁴ = 28 …(1)

a₂ + a₄ = 29 ⇒ ar + ar³ = 29 

⇒ ar(1 + r²) = 29 

a² r² (1 + r²)² = (29)² …(2) 

By Eq. (1) & (2)

$\frac{\mathrm{r}^{2}}{\left(1+\mathrm{r}^{2}\right)^{2}}=\frac{28}{29 \times 29}$

⇒ $\frac{\mathrm{r}}{1+\mathrm{r}^{2}}=\frac{\sqrt{28}}{29} \Rightarrow \mathrm{r}=\sqrt{28}$

∵ a² r⁴ = 28 ⇒ a² × (28)² = 28 

⇒ $a=\frac{1}{\sqrt{28}}$

∴ $a_{6}=a r^{5}=\frac{1}{\sqrt{28}} \times(28)^{2} \sqrt{28}=784$

Hence option 3 is correct

Calculation

P(2λ + 1, 3λ + 2, 4λ + 3) on L₁

Q(3µ + 2, 4µ + 4, 5µ + 5) on L₂

Dr’s of PQ = 3µ – 2λ + 1, 4µ – 3λ + 2, 5µ – 4λ + 2

PQ ⊥ L₁ 

⇒ (3µ – 2λ + 1)2 + (4µ – 3λ + 2)3 + (5µ – 4λ + 2)4 = 0

38µ – 29λ + 16 = 0 …(1)

PQ ⊥ L₂ 

⇒ (3µ – 2λ + 1)3 + (4µ – 3λ + 2)4 + (5µ – 4λ + 2)5 = 0

50µ – 38λ + 21 = 0 …(2)

By (1) & (2) 

$λ=\frac{1}{3} ; \mu=\frac{-1}{6}$

∴ \mathrm{P}\left(\frac{5}{3}, 3, \frac{13}{3}\right) & \ \mathrm{Q}\left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right)

Line PQ 

$\begin{array}{ccc} \frac{x-\frac{5}{3}}{\frac{1}{6}} & \frac{y-3}{\frac{-1}{3}} & \frac{z-\frac{13}{3}}{\frac{1}{6}} \end{array}$

$\frac{x-\frac{5}{3}}{1}=\frac{y-3}{-2}=\frac{z-\frac{13}{3}}{1}$

$\text { Point }\left(\frac{14}{3},-3, \frac{22}{3}\right)$

lies on the line PQ 

Hence option 4 is correct

Calculation

$\rm \mathrm{e}^{5(ℓ n x)^{2}+3}=\mathrm{x}^{8}$

⇒ $\ln \mathrm{e}^{5(ℓ \mathrm{n} x)^{2}+3}=\ln \mathrm{x}^{8}$

⇒ 5(ℓnx)² + 3 = 8ℓnx

(ℓnx = t)

⇒ 5t² – 8t + 3 = 0

\(\mathrm{t}{1}+\mathrm{t}{2}=\frac{8}{5}\)

\(\ell \operatorname{nx}{1} x{2}=\frac{8}{5}\)

x₁ x₂ = e^8/5

Hence option 1 is correct

Calculation

T_n = S_n – S_n–1

⇒ $\mathrm{T}_{\mathrm{n}}=\frac{1}{8}(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3)$

⇒ $\frac{1}{T_{n}}=\frac{8}{(2 n-1)(2 n+1)(2 n+3)}$

\(\lim {n \rightarrow \infty} \sum{r=1}^{n} \frac{1}{T_{r}}=\lim {n \rightarrow \infty} 8 \sum{r=1}^{n} \frac{1}{(2 n-1)(2 n+1)(2 n+3)}\)

= $\lim _{n \rightarrow \infty} \frac{8}{4} \sum\left(\frac{1}{(2 n-1)(2 n+1)}-\frac{1}{(2 n+1)(2 n+3)}\right)$

= $\lim _{\mathrm{n} \rightarrow \infty} 2\left[\left(\frac{1}{1.3}-\frac{1}{3.5}\right)+\left(\frac{1}{3.5}-\frac{1}{5.7}\right)+\ldots\right]$

= $\frac{2}{3}$

Hence option 3 is correct

Calculation 

= \(\underbrace{12 \mathrm{C}{2}}{\substack{\text { Sclection of two } \ \text { letters before } \mathrm{M}}} \times \underbrace{{ }^{13} \mathrm{C}{2}}{\substack{\text { Selection of two } \ \text { letters after } \mathrm{M}}}=5148\)

Hence option 4 is correct

Calculation

$\frac{\mathrm{dx}}{\mathrm{dy}}+\left(\frac{1}{\mathrm{y}^{2}}\right) \mathrm{x}=\frac{1}{\mathrm{y}^{3}}$

$\text { I.F. }=\mathrm{e}^{\int \frac{1}{{y}^{2}} \mathrm{dy}}=\mathrm{e}^{-\frac{1}{y}}$

⇒ $x \cdot e^{-\frac{1}{y}}=\int\left(e^{-\frac{1}{y}}\right) \cdot \frac{1}{y^{3}} d y$

Put $-\frac{1}{y}=t$

$+\frac{1}{\mathrm{y}^{2}} \mathrm{dy}=\mathrm{dt}$

$x \cdot e^{-\frac{1}{y}}=-\int t \cdot e^{t} d t$

$x \cdot e^{-\frac{1}{y}}=-t e^{t}+e^{t}+C$

$x \cdot e^{-\frac{1}{y}}=\frac{+1}{y} e^{-\frac{1}{y}}+e^{-\frac{1}{y}}+C$

x = 1, y = 1 

$\frac{1}{\mathrm{e}}=\frac{1}{\mathrm{e}}+\frac{1}{\mathrm{e}}+\mathrm{C}$

⇒ $\mathrm{C}=-\frac{1}{\mathrm{e}}$

Put y = $\frac{1}{2}$

$\frac{\mathrm{x}}{\mathrm{e}^{2}}=\frac{2}{\mathrm{e}^{2}}+\frac{1}{\mathrm{e}^{2}}-\frac{1}{\mathrm{e}}$

x = 3 – e

Hence option 3 is correct

Calculation

y = x² + px – 3

Let P(α, 0), Q(β, 0), R(0, –3)

Circle with centre (–1, –1) is (x + 1)² + (y + 1)² = r²

Passes through (0, –3)

1² + (–2)² = r²]

r² = 5

(x + 1)² + (y + 1)² = 5

Put y = 0

(x + 1)² = 5 – 1

(x + 1)² = 4

x + 1 = ±2

x = 1 or x = –3 

P(1, 0) and Q(–3, 0)

Area of ΔPQR = $\frac{1}{2}\left|\begin{array}{ccc} 1 & 0 & 1 \ -3 & 0 & 1 \ 0 & -3 & 1 \end{array}\right|=6$

Hence option 2 is correct

Calculation. 

S₁ : (x + 2)² + (y – 2)² = 2²

S₂ : (x – 2)² + (y – 5)² = r²

Both circle intersect at two points 

∴ |r₁ – r₂| < c₁ c₂ < r₁ + r₂

|r – 2| < 5 < 2 + r 

⇒ 3 < r < 7

r ∈ (3, 7)

α = 3, β = 7

3β – 2α = 15

Hence option 1 is correct

Calculation 

f(x) = (7tan⁶ x – 3tan² x)(sec² x)

$I_{1}=\int_{0}^{\pi / 4}\left(7 \tan ^{6} x-3 \tan ^{2} x\right)\left(\sec ^{2} x\right) d x$

Put tanx = t

$I_{1}=\int_{0}^{1}\left(7 t^{6}-3 t^{2}\right) d t=\left[t^{7}-t^{3}\right]_{0}^{1}=0$

$I_{2}=\int_{0}^{\pi / 4} x \underbrace{\left(7 \tan ^{6} x-3 \tan ^{2} x\right)\left(\sec ^{2} x\right)}_{\mathrm{II}} d x$

= \(\left[x\left(\tan ^{7} x-\tan ^{3} x\right)\right]{0}^{\pi / 4}-\int{0}^{\pi / 4}\left(\tan ^{7} x-\tan ^{3} x\right) d x\)

= $0-\int_{0}^{\pi / 4} \tan ^{3} x\left(\tan ^{2} x-1\right)\left(1+\tan ^{2} x\right) d x$

Put tanx = t

= $-\int_{0}^{1}\left(t^{5}-t^{3}\right) d t=-\left[\frac{t^{6}}{6}-\frac{t^{4}}{4}\right]=\frac{1}{12}$

7I₁ + 12I₂ = 1 

Hence option 3 is correct

Calculation 

f(x + y) = f(x) f′(y) + f′(x) f(x)

Put = x = y = 0

f(0) = f(0)f′(0) + f′(0)f(0) 

f′(0) = $\frac{1}{2}$

Put y = 0

f(x) = f(x) f′(0) + f′(x)f(0)

$f(x)=\frac{1}{2} f(x)+f^{\prime}(x)$

$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{f}(\mathrm{x})}{2}$

$\frac{d y}{d x}=\frac{y}{2} \Rightarrow \int \frac{d y}{y}=\int \frac{d x}{2}$

⇒ $\ell n y=\frac{x}{2}+c$

∵ f (0) = 1 ⇒ C = 0 

$\ell \mathrm{ny}=\frac{\pi}{2} \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x} / 2}$

$\ell n \ \mathrm{f}(\mathrm{n})=\frac{\mathrm{n}}{2}$

$\sum_{\mathrm{n}=1}^{100} \ell \mathrm{f}(\mathrm{n})=\frac{1}{2} \sum_{\mathrm{n}=1}^{100} \mathrm{n}=\frac{5050}{2}$

= 2525

Hence option 2 is correct

Concept:

Counting reduced proper fractions:

• Set A: A = {1, 2, 3, …, 10}.
• Set B: B = {m/n : m,n ∈ A, m < n, gcd(m,n)=1}.
• Key idea: For a fixed denominator n, the number of integers
• 1 ≤ m < n with gcd(m,n)=1 equals φ(n).
• Important terms: Euler's totient function φ(n),
• reduced fraction, distinct fractions.
• Therefore: n(B) = Σ_n=2¹⁰ φ(n).

Calculation:

A = {1, 2, ….10}

\mathrm{B}\left{\frac{\mathrm{~m}}{\mathrm{n}}=\mathrm{m}, \mathrm{n} \in \mathrm{~A}, \mathrm{~m}<\mathrm{n}, \operatorname{gcd}(\mathrm{~m}, \mathrm{n})=1\right}

n(B)

\mathrm{n}=2 \quad\left{\frac{1}{2}\right}

⇒ φ(2) = 1

\mathrm{n}=3 \quad\left{\frac{1}{3}, \frac{2}{3}\right}

⇒ φ(3) = 2

\mathrm{n}=4 \quad\left{\frac{1}{4}, \frac{3}{4}\right}

⇒ φ(4) = 2

\mathrm{n}=5 \quad\left{\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}\right}

⇒ φ(5) = 4

\mathrm{n}=6 \quad\left{\frac{1}{6}, \frac{5}{6}\right}

⇒ φ(6) = 2

\mathrm{n}=7 \quad\left{\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\right}

⇒ φ(7) = 6

\mathrm{n}=8 \quad\left{\frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}\right}

⇒ φ(8) = 4

\mathrm{n}=9 \quad\left{\frac{1}{9}, \frac{2}{9}, \frac{4}{9}, \frac{5}{9}, \frac{7}{9}, \frac{8}{9}\right}  

⇒ φ(9) = 6 

\mathrm{n}=10 \quad\left{\frac{1}{10}, \frac{3}{10}, \frac{7}{10}, \frac{9}{10}\right}

⇒ φ(10) = 4

Hence n(B) = 31

∴ n(B) = 31.

Calculation

$y^{2}=2 \sqrt{3} x$

$(x-2 \sqrt{3})^{2}+y^{2}=(2 \sqrt{3})^{2}$

$A=\frac{π r^{2}}{2}-2 \int_{0}^{2 \sqrt{3}} \sqrt{2 \sqrt{3} x} d x$

$\frac{π(12)}{2}-2 \sqrt{2 \sqrt{3}} \frac{\left(\mathrm{x}^{3 / 2}\right)_{0}^{2 / 3}}{3 / 2}$

= 6π – 16

Hence option 3 is correct

Calculation

$P=\frac{\frac{6}{10} \times \frac{5}{9}}{\frac{4}{10} \times \frac{6}{9}+\frac{6}{10} \times \frac{5}{9}}=\frac{5}{9}$

m = 5, n = 9

m + n = 14

Calculation

be = 13, b = 5

a² = b² (e² – 1)

= b² e² – b²

= 169 – 25 = 144

$\ell(\mathrm{LR})=\frac{2 \mathrm{a}^{2}}{\mathrm{~b}}=\frac{2 \times 144}{5}=\frac{288}{5}$

Hence option 3 is correct

Ans. (34)

Sol.

f(x) is continuous and differentiable

at x = 1; LHL = RHL, LHD = RHD

–3a – 2 = a² + b, –6a = b

a = 2, 1; b = –12 

f(x)=\left{\begin{array}{ccc} -6 x^{2}-2 & ; & x<1 \ 4-12 x & ; & x \geq 1 \end{array}\right.

Area = $\int_{-\sqrt{3}}^{1}\left(-6 x^{2}-2+20\right) d x+\int_{1}^{2}(4-12 x+20) d x$

$16+12 \sqrt{3}+6=22+12 \sqrt{3}$

Calculation

$\int_{0}^{1}(1+x)^{11} d x=\left[C_{0} x+\frac{C_{1} x^{2}}{2}+\frac{C_{2} x^{3}}{3}+\ldots\right]_{0}^{1}$

$\frac{2^{12}-1}{12}=C_{0}+\frac{C_{1}}{2}+\frac{C_{2}}{3}+\frac{C_{3}}{4}+\ldots$

$\int_{-1}^{0}(1+x)^{11} d x=\left[C_{0} x+\frac{C_{1} x^{2}}{2}+\frac{C_{2} x^{3}}{3}+\ldots\right]_{-1}^{0}$

$\frac{1}{12}=C_{0}-\frac{C_{1}}{2}+\frac{C_{2}}{3}-\frac{C_{3}}{4}+\ldots$

$\frac{2^{12}-2}{12}=2\left(\frac{C_{1}}{2}+\frac{C_{3}}{4}+\frac{C_{5}}{6}+\ldots\right)$

\(\frac{\mathrm{C}{1}}{2}+\frac{\mathrm{C}{3}}{4}-\frac{\mathrm{C}_{5}}{6}+\ldots=\frac{2^{11}-1}{12}=\frac{2047}{12}\)

m - n = 2047 - 12 = 2035

Calculation

|A| = –2

det(3adj(–6adj(3A)))

= 33 det(adj(–adj(3A)))

= 3³ (–6)6 (det(3A))⁴

= 3²¹ × 2¹⁰

m + n = 10

mn = 21

m = 7 ; n = 3

4m + 2n = 28 + 6 = 34

Calculation

Point B

(3λ + 1, –λ + 1, –1) ≡ (2µ + 2, 0, αµ – 4)

3λ + 1 = 2µ + 2

–λ + 1 = 0

–1 = αµ – 4

λ = 1, µ = 1, α = 3

B(4, 0, –1)

Let Point ‘P’ is (2δ + 2, 0, 3δ – 4)

Dr’s of AP < 2δ + 1, –1, 3δ – 3 >

$\mathrm{AP} \perp \mathrm{~L}_{2} \Rightarrow δ=\frac{7}{13}$

$\mathrm{P}\left(\frac{40}{13}, 0, \frac{-31}{13}\right)$

$2 \sigma δ(\mathrm{~PB})^{2}=26 \times 3 \times\left(\frac{144}{169}+\frac{324}{169}\right)$

= 216

Calculation

$\overrightarrow{\mathrm{c}}=\left(\frac{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}}{|\overrightarrow{\mathrm{a}}|}\right) \frac{\overrightarrow{\mathrm{a}}}{|\overrightarrow{\mathrm{a}}|}$

⇒ $\left(\frac{\lambda+8}{9}\right)(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$

$|\vec{a}+\vec{c}|-=7 \Rightarrow \lambda=4$

Area of parallelogram

= $|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}|=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ \frac{4}{3} & \frac{8}{3} & \frac{8}{3} \ 4 & 0 & 4 \end{array}\right|$

= 16

CONCEPT:

Faraday's Laws of Electrolysis

• Faraday's first law of electrolysis states that the amount of substance deposited at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.
• The relationship can be expressed as:

m = (ItM)/(nF)

where m is the mass of the substance deposited, I is the current, t is the time, M is the molar mass of the substance, n is the number of electrons transferred per ion, and F is the Faraday constant.

EXPLANATION:

• For the electrolysis of aluminium chloride (AlCl₃), the reaction at the cathode is:

Al³⁺ + 3e^- → Al

• Given data:
• Current (I) = 2A
• Time (t) = 30 minutes = 30 × 60 = 1800 seconds
• Molar mass of Aluminium (M) = 27 g mol^-1
• Faraday constant (F) = 96500 C mol^-1
• In the reaction, the number of electrons (n) transferred per ion is 3.
• Using the formula:

m = (ItM)/(nF)

m = (2 × 1800 × 27)/(3 × 96500)

m = (97200)/(289500)

m = 0.336 g

Therefore, the amount of aluminium deposited at the cathode is 0.336 g.

CONCEPT:

Radioactive Decay

• Radioactive decay is a process by which the nucleus of an unstable atom loses energy by emitting radiation. The decay constant (λ) is a probability rate constant that quantifies the likelihood of decay of an atom per unit time.
• Key characteristics of radioactive decay:
• The decay constant (λ) is independent of temperature.
• The half-life (t_1/2) of a radioactive substance is the time required for half of the radioactive atoms to decay. The relationship between half-life and decay constant is given by:

t_1/2 = ln(2) / λ

EXPLANATION:

• Statement 1: Amount of radioactive substance remained after three half-lives is 1/8 of the original amount.
• This is true. After one half-life, half of the original amount remains. After three half-lives, (1/2)³ = 1/8 of the original amount remains.
• Statement 2: Decay constant does not depend upon temperature.
• This is true. The decay constant is a fundamental property of the radioactive substance and is not affected by temperature.
• Statement 3: Decay constant increases with increase in temperature.
• This is not true. The decay constant is independent of temperature.
• Statement 4: Half life is ln(2) times of 1/rate constant.
• This is true. The half-life (t_1/2) is given by t_1/2 = ln(2) / λ, where λ is the decay constant.

Therefore, the statement that is not true for radioactive decay is: Decay constant increases with increase in temperature.

CONCEPT:

Stereoisomers and Geometrical Isomerism

• Stereoisomers are compounds that have the same molecular formula and connectivity of atoms but differ in the spatial arrangement of atoms.
• Geometrical isomerism occurs in compounds with restricted rotation around a double bond or ring, resulting in cis-trans isomers.
• Each chiral center or double bond contributes to possible isomerism.

EXPLANATION:

It has 4 stereoisomers $\left[\begin{array}{ll} \mathrm{R} \text { cis } & \mathrm{R} \text { trans } \ \mathrm{Scis} & \mathrm{Strans} \end{array}\right]$

• Two factors contribute to stereoisomerism here:
• The double bond between the third and fourth carbon introduces cis and trans isomers.
• The chiral carbon (second carbon with -CH₃, -CH=CH-CH₃, -OH, and -H as substituents) introduces R and S configurations.
• Combining these, the molecule can exhibit the following configurations:
• Cis with R configuration
• Cis with S configuration
• Trans with R configuration
• Trans with S configuration
• Hence, the molecule has 4 possible stereoisomers.

Therefore, the total number of stereoisomers for the given molecule is 4.

EXPLANATION:​.

Electronegativity

• Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. The Pauling scale is the most commonly used scale for electronegativity.
• The electronegativity values (on the Pauling scale) for some elements are as follows:
• | | Li | Be | B | C | N | O | F | | --- | --- | --- | --- | --- | --- | --- | --- | | (E.N.) = On pauling scale | 1 | 1.5 | 2 | 2.5 | 3 | 3.5 | 4.0 |

For elements in the second period and third period:

• | | Na | Mg | Al | Si | P | S | Cl | | --- | --- | --- | --- | --- | --- | --- | --- | | (E.N.) = | 0.9 | 1.2 | 1.5 | 1.8 | 2.1 | 2.5 | 3.0
  |

Therefore, the electronegativity order that is incorrect is: Al < Mg < B < N

CONCEPT:

Electronic Configuration of Lanthanoids

• Lanthanoids are a series of 15 elements from lanthanum (La) to lutetium (Lu) in the periodic table, characterized by the filling of the 4f orbitals.
• The general electron configuration of lanthanoids is [Xe] 4fⁿ 5d^x 6s².
• The common oxidation states of lanthanoids include +2, +3, and sometimes +4.
• Key electron configurations:
• Europium (Eu): [Xe] 4f⁷ 6s²
• Gadolinium (Gd): [Xe] 4f⁷ 5d¹ 6s²
• Terbium (Tb): [Xe] 4f⁹ 6s²
• Samarium (Sm): [Xe] 4f⁶ 6s²

EXPLANATION:

• To determine which ions have a 4f⁷ configuration, we must consider the electron loss when forming the given ions:
• Europium (Eu):
• Eu²⁺: [Xe] 4f⁷ (Loses 2 electrons from 6s²)
• Eu³⁺: [Xe] 4f⁶ (Loses 2 electrons from 6s² and 1 electron from 4f⁷)
• Gadolinium (Gd):
• Gd³⁺: [Xe] 4f⁷ (Loses 2 electrons from 6s² and 1 electron from 5d¹)
• Terbium (Tb):
• Tb³⁺: [Xe] 4f⁸ (Loses 2 electrons from 6s² and 1 electron from 4f⁹)
• Samarium (Sm):
• Sm²⁺: [Xe] 4f⁶ (Loses 2 electrons from 6s²)

Therefore, the lanthanoid ions with a 4f⁷ configuration are Eu²⁺ and Gd³⁺.

EXPLANATION:

Trends in the Periodic Table

• Ionic Radii: The size of the ionic radii decreases as the positive charge of the ion increases.
• Order: Al³⁺ < Mg²⁺ < Na⁺ < F^-
• Ionisation Enthalpy: The energy required to remove an electron from an atom in its gaseous state. It generally increases across a period from left to right.
• Order: B < C < O < N
• Metallic Character: The degree to which an element exhibits metallic properties, which decreases across a period and increases down a group.
• Order: B < Al < Mg < K
• Electronegativity: The tendency of an atom to attract electrons in a chemical bond. It generally increases across a period from left to right.
• Order: Si < P < S < Cl

The correct matching is  A-IV, B-I, C-II, D-III

CONCEPT:

Vitamins and their Chemical Names

• Vitamins are essential nutrients required by the body to function properly. Each vitamin has a specific chemical name.
• Ascorbic acid is the chemical name for Vitamin C.

EXPLANATION:

• Adipic acid: This is not a vitamin, but a dicarboxylic acid commonly used in the production of nylon.
• Aspartic acid: This is not a vitamin, but an amino acid involved in the synthesis of proteins.
• Ascorbic acid: This is Vitamin C, an essential nutrient for the repair of tissues and the enzymatic production of certain neurotransmitters.
• Saccharic acid: This is not a vitamin, but an oxidized derivative of glucose.

Therefore, the acid that is a vitamin is ascorbic acid (Vitamin C).

The correct answer is: (3) Ascorbic acid

CONCEPT:

First Law of Thermodynamics

• The first law of thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (q) plus the work done on the system (w):
• ΔU = q + w
• In a thermally insulated (adiabatic) system, there is no heat exchange with the surroundings, so:
• q = 0
• When work is done on the system (for example, by mechanically stirring the liquid), the work done (w) is positive, which increases the internal energy (ΔU) of the system.

EXPLANATION:

• Since the vessel is thermally insulated, there is no heat exchange with the surroundings:
• q = 0
• The work is done on the liquid by mechanically stirring it, so:
• w > 0
• According to the first law of thermodynamics:
• ΔU = q + w
• Since q = 0, we have ΔU = w
• When w > 0, ΔU > 0

Therefore, the correct option for the thermodynamic parameters is  ΔU > 0, q = 0, w > 0​

CONCEPT:

Bohr's Model of the Atom

• In Bohr's model, the radius of the ( n )th orbit of an atom is given by:
• $r_n = \frac{n^2 a_0}{Z}$
• where:
• ( rn ) = radius of the ( n )th orbit
• ( a0 ) = Bohr radius (radius of the first stationary state of the hydrogen atom)
• ( n ) = principal quantum number
• ( Z ) = atomic number of the element
• For the first excited state, ( n = 2 ).

APPLICATION TO HELIUM ION (He⁺):

• For the Helium ion (He⁺), the atomic number ( Z ) is 2.
• For the first excited state, ( n = 2 ).
• Substituting these values into the formula for the radius:
• $r = \frac{n^2 a_0}{Z} \ = \frac{(2)^2 a_0}{2} \ = \frac{4 a_0}{2} \ = 2 a_0$

Therefore, the radius of the first excited state of the Helium ion is r = 2a0

CONCEPT:

SN₁ and SN₂ Reactions

• SN₁ Mechanism: This is a two-step mechanism where the rate-determining step involves the formation of a carbocation intermediate. It is favored by:
• Stable carbocations (tertiary > secondary > primary).
• Polar protic solvents.
• SN₂ Mechanism: This is a single-step mechanism involving a backside attack by the nucleophile. It is favored by:
• Less steric hindrance (methyl > primary > secondary; tertiary is unfavorable).
• Polar aprotic solvents.

EXPLANATION:

CH3 – O – CH2 – Cl will undergo SN1 mechanism because \(\mathrm{CH}{3}-\mathrm{O}-\stackrel{+}{\mathrm{C}} \mathrm{H}{2}\)  is highly stable. 

• Statement I: CH₃–O–CH₂–Cl undergoes the SN₁ mechanism even though it is a primary halide because the intermediate carbocation (CH₃–O–CH₂⁺) formed is stabilized by resonance with the oxygen atom.
• Statement II: CH₃–C(CH₃)₂–CH₂–Cl (neopentyl chloride) does not undergo the SN₂ mechanism easily due to significant steric hindrance around the reactive center, which obstructs the nucleophilic backside attack.
• Both statements are correct based on the mechanisms and the structural properties of the molecules.

Therefore, both Statement I and Statement II are correct.

CONCEPT:

Reactions of Propyne with Sodium and Sodium Amide

• When alkynes react with metallic sodium in the presence of liquid ammonia, they release hydrogen gas. The number of moles of H₂ gas liberated depends on the available acidic hydrogen atoms in the molecule.
• Reaction with sodium amide (NaNH₂) leads to the formation of a sodium salt of the alkyne and releases ammonia (NH₃) gas. This reaction is used to identify terminal alkynes.

EXPLANATION:

• Statement I: One mole of propyne (CH₃–C≡CH) reacts with an excess of sodium (Na) to liberate 0.5 moles of hydrogen gas (H₂), as each terminal hydrogen in the alkyne contributes to the reaction. This statement is correct.
• Statement II: Four grams of propyne contain 0.1 moles of propyne (molecular weight = 40 g/mol). When reacted with NaNH₂, this amount of propyne will release 0.1 moles of NH₃ gas, which occupies 2240 mL at STP (not 224 mL as stated). Hence, this statement is incorrect.
• $\frac{4}{40}=0.1 \text { mole } \quad \frac{0.1 \mathrm{~mole}}{2240 \mathrm{~mole}}$
• Statement I is correct but Statement II is incorrect

Therefore, Statement I is correct, but Statement II is incorrect.

CONCEPT:

Equilibrium Constant (K_P) for Gas Reactions

• The equilibrium constant, K_P, is calculated using the partial pressures of the gases involved in the reaction at equilibrium.
• For a reaction:

aA(g) + bB(g) ⇌ cC(g) + dD(g)

K_P = (P_C)^c (P_D)^d / (P_A)^a (P_B)^b

where P represents the partial pressure of the gases at equilibrium.

• Pure solids and liquids do not appear in the expression for K_P.

EXPLANATION:

• The given reaction is:

CO₂(g) + C(s) ⇌ 2CO(g)

• CO₂ is the reactant gas.
• Graphite (C) is a solid and does not appear in K_P expression.
• CO is the product gas.
• Initial pressure of CO₂ is 0.5 atm.
• Let 'x' be the pressure change due to the conversion of CO₂ to CO at equilibrium:
• Pressure of CO₂ at equilibrium: (0.5 - x) atm
• Pressure of CO at equilibrium: 2x atm
• $\begin{array}{lc} \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{C}(\mathrm{~s}) & \rightleftharpoons 2 \mathrm{CO}(\mathrm{~g}) \ 0.5 & - \ 0.5-\mathrm{x} & 2 \mathrm{x} \end{array}$
• Given: Total pressure at equilibrium is 0.8 atm:
• (0.5 - x) + 2x = 0.8
• Solving for x: 0.5 + x = 0.8
• x = 0.3
• Equilibrium pressures:
• P_CO2 = 0.5 - 0.3 = 0.2 atm
• P_CO = 2(0.3) = 0.6 atm
• Plugging these values into the K_P expression:
• K_P = (P_CO)² / (P_CO2)
• K_P = (0.6)]² / 0.2
• K_P = 0.36 / 0.2
• K_P = 1.8 atm

Therefore, the correct answer is 1.8 atm

CONCEPT:

IUPAC Nomenclature of Organic Compounds

• The IUPAC name of a compound is determined based on the longest carbon chain containing the principal functional group.
• Priority is given to functional groups in the following order: carboxylic acid (-COOH) > ester (-COOR) > ketone (-C=O) > alcohol (-OH) > amine (-NH₂) > alkene > alkyne > alkane.
• Substituents are named and numbered to give the lowest possible position numbers to the functional groups and substituents.

EXPLANATION:

• The given compound contains two functional groups:
• A carboxylic acid (-COOH) group, which has the highest priority.
• An ester group (-COOCH₃) as a substituent.
• The longest carbon chain containing the carboxylic acid group has six carbons (hexane). The main chain is numbered from the carboxylic acid carbon to give substituents the lowest possible numbers.
• The substituents on the chain are:
• A methoxycarbonyl (-COOCH₃) group at position 6.
• Methyl groups at positions 2 and 5.
• The compound's name is written as: 6-Methoxycarbonyl-2,5-dimethylhexanoic acid.

Therefore, the IUPAC name of the compound is: 6-Methoxycarbonyl-2,5-dimethylhexanoic acid.

CONCEPT:

Electrolysis of Sulfuric Acid

• During electrolysis, an electrolyte is decomposed into its constituent ions through the application of an electrical current.
• H₂S₂O₈ (peroxodisulfuric acid) can be produced by the electrolysis of concentrated sulfuric acid solutions.
• At the anode, the half-reaction is:

2 HSO₄^- → H₂S₂O₈ + 2e^-

EXPLANATION:

• To efficiently produce H₂S₂O₈, a concentrated solution of sulfuric acid is required because:
• In a concentrated solution, the concentration of HSO₄^- ions is high enough to facilitate the formation of peroxodisulfuric acid at the anode.
• Dilute solutions of sulfuric acid or sodium sulfate do not provide the necessary ion concentrations for this reaction.

Therefore, the correct answer is: (3) Concentrated solution of sulfuric acid.

CONCEPT:

Fehling's Test

• Fehling's test is used to detect the presence of reducing sugars and aldehydes.
• Compounds with an aldehyde group (-CHO) or α-hydroxy ketones give a positive Fehling's test, resulting in the formation of a red precipitate of cuprous oxide (Cu₂O).
• Ketones and aromatic aldehydes (like benzaldehyde) generally do not give a positive Fehling's test.

EXPLANATION:

• Compound (A): Benzaldehyde (C₆H₅CHO) is an aromatic aldehyde and does not give a positive Fehling's test.
• Compound (B): Acetophenone (C₆H₅COCH₃) is a ketone and does not give a positive Fehling's test.
• Compound (C): Glucose derivative (HOCH₂–CO–(CHOH)₃–CH₂OH) contains an α-hydroxy aldehyde functional group and gives a positive Fehling's test.
• Compound (D): Propionaldehyde (CH₃CH₂CHO) is an aliphatic aldehyde and gives a positive Fehling's test.
• Compound (E): Phenylacetaldehyde (C₆H₅CH₂CHO) is an aliphatic aldehyde and gives a positive Fehling's test.

Therefore, the compounds (C), (D), and (E) give a positive Fehling's test.

CONCEPT:

Crystal Field Stabilization Energy (CFSE)

• CFSE is the energy difference between the energy of an electron configuration in the ligand field and in an isotropic field.
• It depends on:
• The splitting of d-orbitals (t_2g and e_g) in the crystal field.
• The number of electrons in these orbitals.
• CFSE is calculated as:

CFSE = [(number of electrons in t_2g) × (-0.4Δ₀)] + [(number of electrons in e_g) × (+0.6Δ₀)]

EXPLANATION:

For complex K3 [Fe(SCN)6] 

Calculation of CFSE

= (–0.4 × 3 + 0.6 × 2) Δ0

= 0 Δ0

• [Fe(NH₃)₆]Br₂:
• Fe²⁺ (d⁶) forms a high-spin complex due to NH₃ being a weak field ligand. This results in CFSE ≠ 0.
• [Fe(en)₃]Cl₃:
• Fe³⁺ (d⁵) forms a low-spin complex due to en (ethylenediamine) being a strong field ligand. CFSE ≠ 0.
• K₄[Fe(CN)₆]:
• Fe²⁺ (d⁶) forms a low-spin complex due to CN^- being a strong field ligand. CFSE ≠ 0.
• K₃[Fe(SCN)₆]:
• SCN^- is a weak field ligand, and Fe³⁺ (d⁵) forms a high-spin complex. The electron configuration is t_2g³ e_g², resulting in:
• CFSE = [-0.4(3) + 0.6(2)]Δ₀
• CFSE = 0

Therefore, in K₃[Fe(SCN)₆], the CFSE (Δ₀) is equal to zero.

CONCEPT:

Boiling Point Elevation

• The boiling point elevation (ΔT_b) of a solution is directly proportional to the molal concentration (m) of the solute and the van't Hoff factor (i).
• The formula is given by:

ΔT_b = i × K_b × m

where K_b is the ebullioscopic constant of the solvent.

• For ionic compounds, the van't Hoff factor (i) represents the number of particles the compound dissociates into in solution.
• For non-electrolytes like urea, i = 1.

EXPLANATION:

ΔTb = i Kb . m . ∝ i.C.

where C = concentration 

B.P. order (ii) < (i) < (iii) < (iv)

Therefore, the correct answer is (ii) < (i) < (iii) < (iv).

CONCEPT:

Aromatic Substitution and Diazotization

• Bromination in AcOH: Bromine in acetic acid selectively introduces a bromine atom at the para position relative to the nitro group (-NO₂) due to its electron-withdrawing nature.
• Reduction: The nitro group (-NO₂) is reduced to an amino group (-NH₂) using Sn and HCl.
• Diazotization: Amino groups react with NaNO₂ and HCl at low temperatures (273 K) to form diazonium salts (-N₂⁺).
• Reaction with Ethanol: Diazonium salts react with ethanol (EtOH) to form aldehydes, specifically acetaldehyde (CH₃CHO), as a product of the Etard reaction mechanism.

EXPLANATION:

• Step 1: Bromination
• The nitrobenzene reacts with Br₂ in acetic acid (AcOH) to produce para-bromo nitrobenzene.
• Step 2: Reduction
• Para-bromo nitrobenzene is reduced with Sn and HCl to form para-bromo aniline (p-bromoaniline).
• Step 3: Diazotization
• p-Bromoaniline reacts with NaNO₂ and HCl at 273 K to form para-bromo benzene diazonium chloride.
• Step 4: Reaction with Ethanol
• The diazonium salt reacts with ethanol (EtOH), forming para-bromo benzene and acetaldehyde (CH₃CHO).

Therefore, the products formed are para-bromo benzene and acetaldehyde (CH₃CHO).

CONCEPT:

Magnetic Behaviour and Geometries

• The magnetic behaviour of a complex can often be used to infer its geometry and the oxidation state of the central metal ion.
• [NiCl₄]²⁻:
• Ni²⁺ has the electronic configuration [Ar] 3d⁸.
• In a tetrahedral field, the 3d orbitals do not experience significant splitting, retaining unpaired electrons and leading to a paramagnetic nature.
• This suggests a tetrahedral geometry.
• [Ni(CO)₄]:
• Ni in zero oxidation state (Ni(0)) has the electronic configuration [Ar] 3d¹⁰.
• In a tetrahedral field, the d orbitals become fully paired in the presence of strong field ligands (like CO), leading to a diamagnetic nature.
• This suggests a tetrahedral geometry.

EXPLANATION:

• [NiCl₄]²⁻:
• Oxidation state of Ni is +2.
• Ni²⁺ electronic configuration: [Ar] 3d⁸.
• If the geometry is tetrahedral, the hybridization is sp³.
• With 2 unpaired electrons, it exhibits paramagnetism.
• [Ni(CO)₄]:
• Oxidation state of Ni is 0.
• Ni(0) electronic configuration: [Ar] 3d¹⁰.
• If the geometry is tetrahedral, the hybridization is sp³.
• With no unpaired electrons, it exhibits diamagnetism.

Therefore, the correct answer is [NiCl₄]²⁻: Ni²⁺, tetrahedral, [Ni(CO)₄]: Ni(0), tetrahedral

CONCEPT:

Geometrical Isomerism

• Geometrical isomerism (cis-trans isomerism) occurs in compounds with restricted rotation around a double bond or in cyclic structures.
• Cis-isomer: Identical groups are on the same side of the double bond.
• Trans-isomer: Identical groups are on opposite sides of the double bond.
• Properties such as dipole moment and melting point differ for cis and trans isomers:
• Cis-isomers usually have higher dipole moments due to group orientation.
• Trans-isomers generally have higher melting points due to better packing in the solid state.

EXPLANATION:

• (A) Propene does not show geometrical isomerism:
• Geometrical isomerism requires two different groups on each carbon of the double bond, which propene (CH₃–CH=CH₂) lacks.

Incorrect

• (B) Trans-isomer has identical groups on opposite sides:
• This is true; geometrical isomers differ based on the position of groups around the double bond.

Correct

• (C) Cis-but-2-ene has a higher dipole moment than trans-but-2-ene:

(dipole moment only)

• In cis-isomer, the dipoles of substituent groups add up, while in trans-isomer, the dipoles cancel out.

Correct

• (D) 2-Methylbut-2-ene shows two geometrical isomers:

 (does not show GI) 

• This is incorrect as 2-methylbut-2-ene does not have two different groups on each carbon of the double bond.

Incorrect

• (E) Trans-isomer has a lower melting point than cis-isomer:

 (Melting point) 

• Trans-isomers generally have a higher melting point due to better packing in the solid state.

Incorrect

Therefore, the incorrect statements are: (A), (D), and (E).

CONCEPT:

• The volume of CO₂ gas can be determined using the moles of CO₂ gas in the sealed container.
• The number of moles of CO₂ is derived from the neutralization of the unreacted excess Ca(OH)₂ with HCl.

EXPLANATION:

• Let moles of CO₂ be n .
• Since the entire amount of CO₂ reacts with exactly half the initial amount of Ca(OH)₂, the initial moles of Ca(OH)₂ must be 2n .
• The excess Ca(OH)₂ remaining would then be n moles.
• The gram equivalents of Ca(OH)₂ remaining must equal the gram equivalents of HCl used to neutralize it.
• n x 2 = 0.1 x 40/1000 $\times 1$

2n = 0.004

n = 0.002

• Setting up the equation:
• This simplifies to:

The moles of CO₂ gas is 0.002 moles.

Using the molar volume of an ideal gas (22,400 cm³ at STP):

$0.002 \times 22,400 = 44.8 \text{ cm}^3$

• Volume of CO₂:
• Nearest integer: 45 cm³

Therefore, the volume of CO₂ is 45 cm³

CONCEPT:

Carius Method for Estimation of Halogens

• In this method, an organic compound containing halogens is treated with silver nitrate to form silver halide precipitate.
• The mass of silver halide is used to determine the amount of halogen present in the organic compound.

EXPLANATION:

• Mass of organic compound = 180 mg = 0.18 g
• Mass of AgCl produced = 143.5 mg = 0.1435 g
• Molar mass of Ag = 108 g/mol
• Molar mass of Cl = 35.5 g/mol
• Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol

To find the moles of AgCl produced:

• n(AgCl) = mass / molar mass
• n(AgCl) = 0.1435 g / 143.5 g/mol
• n(AgCl) = 0.001 mol

Since 1 mol of AgCl contains 1 mole of Cl:

• n(Cl) = n(AgCl) = 0.001 mol

Mass of Cl in the compound:

• mass(Cl) = n(Cl) × molar mass of Cl
• mass(Cl) = 0.001 mol × 35.5 g/mol
• mass(Cl) = 0.0355 g = 35.5 mg

Percentage of Cl in the compound:

• %Cl = (mass(Cl) / mass of organic compound) × 100
• %Cl = (35.5 mg / 180 mg) × 100
• %Cl ≈ 19.72%

Rounding to the nearest specified range:

• The percentage composition of chlorine in the compound is 20%

CONCEPT:

Linear Geometry

• Linear geometry occurs when the bond angle between atoms is 180°.
• Linear molecules or ions are typically formed when:
• The central atom has no lone pairs and is surrounded by two bonding domains (sp hybridization).
• The central atom has lone pairs but achieves a linear shape due to symmetric electron distribution (e.g., sp³d hybridization).

EXPLANATION:

• BeCl₂: Linear geometry due to sp hybridization (Cl–Be–Cl).

• CO₂: Linear geometry due to sp hybridization (O=C=O).

• N₃⁻: Linear geometry due to sp hybridization (N=N=N).

• XeF₂: Linear geometry due to sp³d hybridization (F–Xe–F, with three lone pairs in equatorial positions).

• NO₂⁺: Linear geometry due to sp hybridization (O=N=O).

• I₃⁻: Linear geometry due to sp³d hybridization (I–I–I, with three lone pairs in equatorial positions).

• Other species such as SO₂, NO₂, and O₃ have bent geometries.

Therefore, the number of species with linear geometry is 6: BeCl₂, CO₂, N₃⁻, XeF₂, NO₂⁺, and I₃⁻.

Ans. (69)  

Sol.

$t_{1 / 2}=\frac{0.693}{K}$​

K = Ae^–Ea/RT

= $10^{20} \times \mathrm{e}^{-\frac{191.48 \times 10^{3}}{8.314 \times 1000}}$

= 10²⁰ × e^–23.031 = 10²⁰ × – e^{ln 10 × 10}

= $\frac{10^{20}}{10^{10}}=10^{10} \mathrm{sec} .$

$\mathrm{t}_{1 / 2}=\frac{0.693}{10^{10}}=6.93 \times 10^{-11}$

= 69.3 × 10^–12 sec

CONCEPT:

First Order Kinetics

• The time required for half of the reactant to convert to product (half-life, t_1/2) can be determined using the rate constant (k).
• The Arrhenius equation relates the rate constant to the temperature and activation energy: $k = A e^{-\frac{E_a}{RT}}$

EXPLANATION:

• Given:
• Activation energy, E_a = 191.48 kJ/mol = 191,480 J/mol
• Frequency factor, A = 10²⁰ s⁻¹
• Temperature, T = 1000 K
• Universal gas constant, R = 8.314 J K⁻¹ mol⁻¹

We first solve for the rate constant (k) using the Arrhenius equation:

• $k = A e^{-\frac{E_a}{RT}} \ k = 10^{20} e^{-\frac{191,480}{8.314 \cdot 1000}} \ k = 10^{20} e^{-\frac{191.48}{8.314}} \ k = 10^{20} e^{-23.031} \ e^{-23.031} \approx 10^{-10} (since ln(10) \approx 2.303 , result closely matches 10^{-10} ) \ k = 10^{20} \cdot 10^{-10} \ k = 10^{10} \text{ s}^{-1}$