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JEE Advanced · Chemistry · Chemical Kinetics

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1
For a first-order reaction R→PR\to P, the following graphs (each plotted against time tt, unless stated otherwise) are shown: (i) [P][P] vs tt, drawn as a concave-up increasing curve; (ii) d[R]dt\dfrac{d[R]}{dt} vs [R][R], drawn as a straight line through the origin with positive slope; (iii) d[P]dt\dfrac{d[P]}{dt} vs tt, drawn as a curve decreasing exponentially from k[R]0k[R]_0 towards zero; (iv) the rate constant kk vs tt, drawn as a horizontal straight line. Which of the given graphs correctly represent(s) the first-order kinetics of R→PR\to P?
Not attempted
A
Graph (i): [P][P] vs tt
t [P] 0
B
Graph (ii): d[R]dt\dfrac{d[R]}{dt} vs [R][R]
[R] d[R]/dt 0
C
Graph (iii): d[P]dt\dfrac{d[P]}{dt} vs tt
d[P]/dt k[R]₀ t
D
Graph (iv): kk vs tt
t k 0
2
For a reversible reaction R⇌PR\rightleftharpoons P at constant temperature, both forward and backward reactions are first-order elementary reactions with rate constants kfk_f and kbk_b respectively. At t=0t=0, [R]=[R]0[R]=[R]_0 and [P]=0[P]=0. If kb=4kfk_b=4k_f, the correct graphical representation of [R]/[R]0[R]/[R]_0 and [P]/[R]0[P]/[R]_0 versus time is
Not attempted
A
Both [R]/[R]0[R]/[R]_0 and [P]/[R]0[P]/[R]_0 approach 0.50.5 (equal at equilibrium)
B
[P]/[R]0[P]/[R]_0 rises to a plateau of 0.80.8 while [R]/[R]0[R]/[R]_0 falls to a plateau of 0.20.2
C
A third graphical pattern (see original figure)
D
A fourth graphical pattern (see original figure)
3
Consider a reaction A+R→ProductA + R \rightarrow \text{Product}. The rate of this reaction is measured to be k[A][R]k[A][R]. At the start of the reaction, the concentration of RR, [R]0[R]_0, is 10 times the concentration of AA, [A]0[A]_0. The reaction can be considered to be a pseudo first order reaction with assumption that k[R]=k′k[R] = k' is constant. Due to this assumption, the relative error (in %) in the rate when this reaction is 40% complete, is ______. [kk and k′k' represent corresponding rate constants]
Not attempted
Your answer: —
Correct answer: 4.17
4
Consider the reaction 2H2(g)+2NO(g)→N2(g)+2H2O(g)2H_2(g) + 2NO(g) \rightarrow N_2(g) + 2H_2O(g), which follows the mechanism: step 1 (fast equilibrium, rate constants k1k_1 forward / k−1k_{-1} reverse): 2NO(g)⇌N2O2(g)2NO(g) \rightleftharpoons N_2O_2(g); step 2 (slow, rate constant k2k_2): N2O2(g)+H2(g)→N2O(g)+H2O(g)N_2O_2(g) + H_2(g) \rightarrow N_2O(g) + H_2O(g); step 3 (fast, rate constant k3k_3): N2O(g)+H2(g)→N2(g)+H2O(g)N_2O(g) + H_2(g) \rightarrow N_2(g) + H_2O(g). The order of the reaction is ______.
Not attempted
Your answer: —
Correct answer: 3
5
The plot of log⁡kf\log k_f versus 1/T1/T for a reversible reaction A(g)⇌P(g)A(g)\rightleftharpoons P(g) is shown (the line passes through log⁡kf=9\log k_f=9 at 1/T=0.002 K−11/T=0.002\ K^{-1}, i.e. at T=500 KT=500\ K). Pre-exponential factors for the forward and backward reactions are 1015 s−110^{15}\ s^{-1} and 1011 s−110^{11}\ s^{-1}, respectively. If the value of log⁡K\log K for the reaction at 500 K500\ K is 66, the value of ∣log⁡kb∣|\log k_b| at 250 K250\ K is ___. [KK = equilibrium constant of the reaction, kfk_f = rate constant of forward reaction, kbk_b = rate constant of backward reaction]
Not attempted
Your answer: —
Correct answer: 5
6
Match the rate expressions in LIST-I for the decomposition of XX with the corresponding profiles provided in LIST-II. XsX_s and kk are constants having appropriate units. LIST-I (I) rate=k[X]Xs+[X]rate = \dfrac{k[X]}{X_s+[X]}, under all possible initial concentrations of XX (II) rate=k[X]Xs+[X]rate = \dfrac{k[X]}{X_s+[X]}, where initial concentrations of XX are much less than XsX_s (III) rate=k[X]Xs+[X]rate = \dfrac{k[X]}{X_s+[X]}, where initial concentrations of XX are much higher than XsX_s (IV) rate=k[X]2Xs+[X]rate = \dfrac{k[X]^2}{X_s+[X]}, where initial concentration of XX is much higher than XsX_s LIST-II (graphs P, Q, R, S, T are shown in the image above) (A) I→\rightarrowP; II→\rightarrowQ; III→\rightarrowS; IV→\rightarrowT (B) I→\rightarrowR; II→\rightarrowS; III→\rightarrowS; IV→\rightarrowT (C) I→\rightarrowP; II→\rightarrowQ; III→\rightarrowQ; IV→\rightarrowR (D) I→\rightarrowR; II→\rightarrowS; III→\rightarrowQ; IV→\rightarrowR
Not attempted
P Q R S T
A
I→\rightarrowP; II→\rightarrowQ; III→\rightarrowS; IV→\rightarrowT
B
I→\rightarrowR; II→\rightarrowS; III→\rightarrowS; IV→\rightarrowT
C
I→\rightarrowP; II→\rightarrowQ; III→\rightarrowQ; IV→\rightarrowR
D
I→\rightarrowR; II→\rightarrowS; III→\rightarrowQ; IV→\rightarrowR
7
For the following reaction 2X+Y→kP2X + Y \xrightarrow{k} P, the rate of reaction is d[P]dt=k[X]\dfrac{d[P]}{dt} = k[X]. Two moles of XX are mixed with one mole of YY to make 1.01.0 L of solution. At 5050 s, 0.50.5 mole of YY is left in the reaction mixture. The correct statement(s) about the reaction is(are) (Use: ln⁡2=0.693\ln 2 = 0.693)
Not attempted
A
The rate constant, kk, of the reaction is 13.86×10−4 s−113.86\times10^{-4}\ s^{-1}.
B
Half-life of XX is 5050 s.
C
At 5050 s, −d[X]dt=13.86×10−3 mol L−1 s−1-\dfrac{d[X]}{dt} = 13.86\times10^{-3}\ mol\ L^{-1}\ s^{-1}.
D
At 100100 s, −d[Y]dt=3.46×10−3 mol L−1 s−1-\dfrac{d[Y]}{dt} = 3.46\times10^{-3}\ mol\ L^{-1}\ s^{-1}.
8
Which of the following plots is(are) correct for the given reaction shown below? ([P]0[P]_0 is the initial concentration of P\textbf{P}.)
Not attempted
(A) (B) (C) (D)
A
Plot (A): t1/2t_{1/2} vs. [P]0[P]_0 is a horizontal line.
B
Plot (B): Initial rate vs. [P]0[P]_0 is a horizontal line.
C
Plot (C): [Q][P]0\frac{[Q]}{[P]_0} vs. time rises with increasing (concave-up) slope.
D
Plot (D): ln⁡([P][P]0)\ln\left(\frac{[P]}{[P]_0}\right) vs. time is a straight line of positive slope through the origin.
9
The decomposition reaction 2N2O5(g)→Δ2N2O4(g)+O2(g)2N_2O_5(g) \xrightarrow{\Delta} 2N_2O_4(g) + O_2(g) is started in a closed cylinder under isothermal isochoric condition at an initial pressure of 1 atm. After Y×103 sY \times 10^3\ s, the pressure inside the cylinder is found to be 1.45 atm1.45\ atm. If the rate constant of the reaction is 5×10−4 s−15\times10^{-4}\ s^{-1}, assuming ideal gas behavior, the value of YY is ___
Not attempted
Your answer: —
Correct answer: 2.3
10
Consider the kinetic data given in the following table for the reaction A+B+C→ProductA + B + C \to \text{Product}: Experiment 1: [A]=0.2[A]=0.2, [B]=0.1[B]=0.1, [C]=0.1[C]=0.1 (mol dm−3^{-3}), Rate =6.0×10−5= 6.0 \times 10^{-5} mol dm−3^{-3} s−1^{-1} Experiment 2: [A]=0.2[A]=0.2, [B]=0.2[B]=0.2, [C]=0.1[C]=0.1, Rate =6.0×10−5= 6.0 \times 10^{-5} Experiment 3: [A]=0.2[A]=0.2, [B]=0.1[B]=0.1, [C]=0.2[C]=0.2, Rate =1.2×10−4= 1.2 \times 10^{-4} Experiment 4: [A]=0.3[A]=0.3, [B]=0.1[B]=0.1, [C]=0.1[C]=0.1, Rate =9.0×10−5= 9.0 \times 10^{-5} The rate of the reaction for [A]=0.15[A] = 0.15, [B]=0.25[B] = 0.25, [C]=0.15[C] = 0.15 mol dm−3^{-3} is found to be Y×10−5Y \times 10^{-5} mol dm−3^{-3} s−1^{-1}. The value of YY is _______
Not attempted
Your answer: —
Correct answer: 6.75
11
Consider the following reversible reaction, A(g)+B(g)⇌AB(g)A(g)+B(g) \rightleftharpoons AB(g). The activation energy of the backward reaction exceeds that of the forward reaction by 2RT2RT (in J mol−1mol^{-1}). If the pre-exponential factor of the forward reaction is 4 times that of the reverse reaction, the absolute value of ΔG∘\Delta G^{\circ} (in J mol−1mol^{-1}) for the reaction at 300 K is ____. (Given: ln⁡(2)=0.7\ln(2) = 0.7, RT=2500RT = 2500 J mol−1mol^{-1} at 300 K and G is the Gibbs energy)
Not attempted
Your answer: —
Correct answer: 8500
12
For a first order reaction A(g)→2B(g)+C(g)A(g) \to 2B(g) + C(g) at constant volume and 300 K, the total pressure at the beginning (t=0t=0) and at time tt are P0P_0 and PtP_t, respectively. Initially, only AA is present with concentration [A]0[A]_0, and t1/3t_{1/3} is the time required for the partial pressure of AA to reach 1/3rd1/3^{rd} of its initial value. The correct option(s) is (are) (Assume that all these gases behave as ideal gases)
Not attempted
A
Graph of ln⁡(3P0−Pt)\ln(3P_0-P_t) vs Time is a straight line with negative slope
B
Graph of t1/3t_{1/3} vs [A]0[A]_0 is a declining straight line
C
Graph of ln⁡(P0−Pt)\ln(P_0-P_t) vs Time is a straight line with negative slope
D
Graph of Rate constant vs [A]0[A]_0 is a horizontal (flat) line
13
In a bimolecular reaction, the steric factor PP was experimentally determined to be 4.5. The correct option(s) among the following is(are)
Not attempted
A
The activation energy of the reaction is unaffected by the value of the steric factor
B
Experimentally determined value of frequency factor is higher than that predicted by Arrhenius equation
C
Since P = 4.5, the reaction will not proceed unless an effective catalyst is used
D
The value of frequency factor predicted by Arrhenius equation is higher than that determined experimentally
14
According to the Arrhenius equation,
Not attempted
A
a high activation energy usually implies a fast reaction.
B
rate constant increases with increase in temperature. This is due to a greater number of collisions whose energy exceeds the activation energy.
C
higher the magnitude of activation energy, stronger is the temperature dependence of the rate constant.
D
the pre-exponential factor is a measure of the rate at which collisions occur, irrespective of their energy.
15
In dilute aqueous H2SO4H_2SO_4, the complex diaquodioxalatoferrate(II) is oxidized by MnO4−MnO_4^-. For this reaction, the ratio of the rate of change of [H+][H^+] to the rate of change of [MnO4−][MnO_4^-] is ______ (single-digit integer, 0-9).
Not attempted
Your answer: —
Correct answer: 8
16
For the elementary reaction M→NM \to N, the rate of disappearance of MM increases by a factor of 8 upon doubling the concentration of MM. The order of the reaction with respect to MM is:
Not attempted
A
4
B
3
C
2
D
1
17
In the reaction, P+Q→R+SP + Q \rightarrow R + S, the time taken for 75% reaction of PP is twice the time taken for 50% reaction of PP. The concentration of QQ varies with reaction time as shown in the figure. The overall order of the reaction is:
Not attempted
A
22
B
33
C
00
D
11
Hint\textbf{Hint}: t75%=2 t50%t_{75\%} = 2\,t_{50\%} is the defining signature of first-order kinetics in PP; a [Q][Q]-vs-time plot that is a straight line falling to zero is the defining signature of zero order in QQ.
Deconstruction\textbf{Deconstruction}: Find the order with respect to PP from the given time-ratio clue, and the order with respect to QQ from the shape of the concentration-time graph, then add them for the overall order. Strategy\textbf{Strategy}: For a first-order reaction, t1/2=ln⁡2kt_{1/2} = \frac{\ln 2}{k} and t75%=ln⁡4k=2ln⁡2k=2 t50%t_{75\%} = \frac{\ln 4}{k} = \frac{2\ln2}{k} = 2\,t_{50\%} - exactly the relationship given, so the reaction is first order in PP. Separately, a species obeying zero-order decay has [Q]=[Q]0−kt[Q] = [Q]_0 - kt, a straight line of negative slope reaching the axis - exactly the graph shown for QQ. Execution\textbf{Execution}: Order in PP = 1 (from the t75%=2t50%t_{75\%}=2t_{50\%} test, which holds for first order and no other simple order). Order in QQ = 0 (linear, not exponential, decay of [Q][Q] with time). Overall order =1+0=1= 1 + 0 = 1. Verification\textbf{Verification}: Check the alternative - if the reaction were second order in PP, t75/t50t_{75}/t_{50} would be 33, not 22; if zero order in PP, the ratio would be 1.51.5. Only first order gives ratio 22, confirming the assignment. Likewise, if QQ had any order >0>0, its concentration-time curve would be exponential/curved, not the straight line shown. Final Answer\textbf{Final Answer}: (D) 11