Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
For reaction $\text{A} \to \text{P}$, rate constant $k = 1.5 \times 10^3$ s$^{-1}$ at 27°C
If activation energy for the above reaction is 60 kJ mol$^{-1}$, then the temperature (in °C) at which rate constant, $k = 4.5 \times 10^3$ s$^{-1}$ is ___________. (Nearest integer)
Given : $\log 2 = 0.30$, $\log 3 = 0.48$, $R = 8.3$ J K$^{-1}$ mol$^{-1}$, $\ln 10 = 2.3$
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Arrhenius Equation - Temperature Dependence of Rate Constant
Consider the reaction $aX \to bY$, for which the rate constant at 30°C is $1 \times 10^{-3}$ mol$^{-1}$ L s$^{-1}$. Which of the following statements are true ?
A. When concentration of 'X' is increased to four times, the rate of reaction becomes 16 times.
B. The reaction is a second order reaction.
C. The half-life period is independent of the concentration of X.
D. Decomposition of $N_2O_5$ is an example of the above reaction.
E. $\ln\frac{[R_0]}{[R]}$ is valid for the above reaction. [Figure: A graph showing ln([R₀]/[R]) vs time with a positive linear relationship]
Choose the correct answer from the options given below :
(1) A and B Only
(2) A, B and C Only
(3)
(4)
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Answer: Option (1)
Key idea: Determining reaction order from rate constant units and applying second-order kinetics properties
Determine order from rate constant units: $k$ has units mol$^{-1}$ L s$^{-1}$ = (mol L$^{-1}$)$^{1-n}$ s$^{-1}$, so $1-n = -1$ giving $n = 2$ (second order)
**Statement A**: For second order, Rate $= k[X]^2$. If $[X] \to 4[X]$, then Rate $\to k(4[X])^2 = 16k[X]^2$ (16 times). TRUE ✓
**Statement B**: From units analysis, reaction is second order. TRUE ✓
**Statement C**: For second order, $t_{1/2} = \frac{1}{k[X]_0}$ depends on initial concentration. FALSE ✗
**Statement D**: Decomposition of $N_2O_5$ is first order, not second order. FALSE ✗
**Statement E**: $\ln\frac{[R_0]}{[R]}$ vs time is linear for first order, not second order (which uses $\frac{1}{[R]}$ vs $t$). FALSE ✗
Only statements A and B are true, so answer is option (1)
Consider the following gas phase reaction being carried out in a closed vessel at 25°C.
$2A(g) \longrightarrow 4B(g) + C(g)$
[Figure: Table showing time (min) vs total pressure of the system (mm Hg): at 30 min, pressure is 300 mm Hg; at $\infty$, pressure is 600 mm Hg]
The pressure of $C(g)$ at 30 minutes time interval would be $\_\_\_\_\_\_\_\_\_\_$ mm Hg. (nearest integer)
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Answer: 20
Key idea: Gas Phase Kinetics - Pressure Analysis using Stoichiometry
For reaction $2A(g) \rightarrow 4B(g) + C(g)$, set up pressure table: Initial $P_0$ for A, at time $t$: $P_A = P_0 - 2x$, $P_B = 4x$, $P_C = x$
Total pressure at any time: $P_{total} = (P_0 - 2x) + 4x + x = P_0 + 3x$
At $t = \infty$, all A reacts: $P_0 - 2x_\infty = 0 \Rightarrow x_\infty = P_0/2$. So $P_0 + 3(P_0/2) = 600$
Solve: $\frac{5P_0}{2} = 600 \Rightarrow P_0 = 240$ mm Hg
At $t = 30$ min: $240 + 3x_{30} = 300 \Rightarrow x_{30} = 20$ mm Hg
Pressure of $C(g)$ at 30 min = $x_{30} = 20$ mm Hg
Consider the given graph showing variation of reactant concentration with time.
Three different reactions were started with identical initial concentration of reactants. Which of the following statement is correct?
[Figure: Graph showing reactant concentration [R] in mol L$^{-1}$ vs time t/s. Three curves labeled 1, 2, and 3 start from initial concentration [R]$_0$ and decrease with time, with curve 1 having the steepest decline, followed by 2, then 3.]
(1) The order of all the three reactions is same.
(2) The rate constant of reaction 3 is larger than the rate constant of reaction 2 if the order of reaction is same for both.
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Answer: Option (1)
Key idea: Relationship between reaction order, rate constant, and concentration-time curves
Observe that all three curves have the same shape/curvature, just different steepness, indicating they follow the same rate law.
For reactions of the same order with different rate constants, the curves are parallel with the same functional form.
First order gas phase reaction
$A \rightarrow B + C$
$p_i$ = initial pressure of gas A, $p_t$ = total pressure of the reaction mixture at time $t$
Expression of rate constant (k) is
(1) $\frac{1}{t}\ln\frac{p_i}{2p_i - p_t}$
(2) $\frac{1}{t}\ln\frac{2p_i}{p_i - p_t}$
(3) $\frac{1}{t}\ln\frac{p_i}{3p_i - 2p_t}$
(4) $\frac{1}{t}\ln\frac{3p_i}{4p_i - p_t}$
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Answer: Option (1)
Key idea: First order kinetics for gas phase reactions with pressure measurements
Set up initial conditions: At $t=0$, pressure of A is $p_i$, B and C are zero, total pressure = $p_i$
At time $t$: Let $x$ = pressure of A reacted. Then pressure of A = $p_i - x$, B = $x$, C = $x$
Total pressure: $p_t = (p_i - x) + x + x = p_i + x$
Solve for pressure of A at time $t$: $x = p_t - p_i$, so pressure of A = $p_i - x = 2p_i - p_t$
Apply first order rate equation: $k = \frac{1}{t}\ln\frac{[A]_0}{[A]_t} = \frac{1}{t}\ln\frac{p_i}{2p_i - p_t}$
If the half life of a first order reaction is 6.93 minutes then the time required for completion of 99% of the reaction will be ________ minutes. (Given : $\log 2 = 0.3010$)
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: First-order reaction kinetics: relationship between half-life, rate constant, and time for percentage completion
For a reaction A $\rightarrow$ P at T K, the half life $(t_{1/2})$ is plotted as a function of initial concentration [A]$_0$ of A as given below.
[Figure: A graph with x-axis labeled "[A]$_0$/mol L$^{-1}$" and y-axis labeled "$t_{1/2}$/s". The graph shows a straight line passing through the origin with positive slope. Two points are marked on x-axis at $1.5 \times 10^{-3}$ and $4 \times 10^{-3}$. A horizontal dashed line at $y = x$ intersects the line, and another horizontal dashed line is shown at $y = 240$.]
The value of $x$ in the given figure is __________ s (Nearest integer)
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Answer: 90
Key idea: Zero-order reaction kinetics: half-life is directly proportional to initial concentration
Observe that $t_{1/2}$ vs $[A]_0$ is a straight line through origin, indicating **zero-order reaction**
For zero-order reaction: $t_{1/2} = \frac{[A]_0}{2k}$, so $t_{1/2} \propto [A]_0$
From graph, at $[A]_0 = 4 \times 10^{-3}$ mol L$^{-1}$, $t_{1/2} = 240$ s
Sucrose hydrolyses in acidic medium into glucose and fructose by first order rate law with $t_{1/2} = 3$ hour. The percentage of sucrose remaining after 6 hours is ______. (Nearest integer)
(Given : $\log 2 = 0.3010$ and $\log 3 = 0.4771$)
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Answer: 25
Key idea: First-order kinetics and half-life relationship
For first-order reaction, use the relationship: $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{3} = 0.231 \text{ hr}^{-1}$
Calculate number of half-lives: $n = \frac{t}{t_{1/2}} = \frac{6}{3} = 2$ half-lives
Decomposition of a hydrocarbon follows the equation $k = (5.5 \times 10^{11} s^{-1}) e^{\frac{-28000K}{T}}$. The activation energy of reaction is __________ kJ mol$^{-1}$. (Nearest Integer)
Given: R = 8.3 J K$^{-1}$ mol$^{-1}$
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Answer: 232
Key idea: Arrhenius Equation and Activation Energy
Write the Arrhenius equation: $k = A e^{-E_a/RT}$
Compare with given equation: $k = (5.5 \times 10^{11}) e^{-28000/T}$
Match the exponents: $\frac{-28000}{T} = \frac{-E_a}{RT}$
Given below are two statements:
$R = 8.314$ J K$^{-1}$ mol$^{-1}$ and 1 cal = 4.2 J
**Statement I:** When Ea = 12.6 kcal/mol, the room temperature rate constant is doubled by a 10°C increase in temperature (298 K to 308 K)
**Statement II:** For a first order reactions $A \to B$,
[Figure: A graph with y-axis labeled $t_{1/2}$/s and x-axis labeled $[A]_0$/mol L$^{-1}$, showing a straight line starting from origin going upward]
Here $[A]_0$ is the initial concentration of A and $t_{1/2}$ is half life of reaction.
In the light of the above statements, choose the **correct** answer from the options given below:
(1) Both **Statement I** and **Statement II** are true
(2) Both **Statement I** and **Statement II** are false
(3) **Statement I** is true but **Statement II** is false
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Answer: Option (3)
Key idea: Arrhenius equation and order of reaction kinetics
**Statement I:** Use Arrhenius equation: $\ln(k_2/k_1) = \frac{E_a}{R}\left[\frac{1}{T_1} - \frac{1}{T_2}\right]$
