JEE Advanced 2018 · Paper 2 · Chemistry Q5

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ChemistryMultiple correct+4 / -2Question 23 of 54

For a first order reaction $\mathrm{A(g)} \to \mathrm{2B(g)} + \mathrm{C(g)}$ at constant volume and 300 K, the total pressure at the beginning ( $t = 0$ ) and at time $t$ are $P_0$ and $P_t$ , respectively. Initially, only A is present with concentration $[\mathrm{A}]_0$ , and $t_{1/3}$ is the time required for the partial pressure of A to reach 1/3 $^{\text{rd}}$ of its initial value. The correct option(s) is (are)

(Assume that all these gases behave as ideal gases)

Plot of ln(3P₀ − P_t) against time: a straight line with negative slope and positive intercept.

Plot of t₁/₃ against [A]₀: a straight line decreasing with [A]₀.

Plot of ln(P₀ − P_t) against time: a straight line with negative slope.

Plot of rate constant against [A]₀: a horizontal straight line.

Answer and worked solution

Official answer: A, D

Pressures. Let $x$ be the pressure of A consumed. Then $p_A=P_0-x$ and $P_t=P_0+2x$ , so
$p_A=\frac{3P_0-P_t}2.$

First order. $\ln p_A=\ln P_0-kt$ , so
$\ln(3P_0-P_t)=\ln2P_0-kt.$
This is a straight line with negative slope against time. (A) is correct.

(B) For first order, $t_{1/3}=\dfrac{\ln3}k$ , independent of $[\ce{A}]_0$ . A decreasing line is wrong.
(C) $P_t>P_0$ , so $P_0-P_t<0$ and its log is undefined. Wrong.
(D) The rate constant does not depend on $[\ce{A}]_0$ , so the plot is a horizontal line. Correct.

(A), (D)

Topics:Integrated rate equations Half-life

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JEE Advanced 2018 · Paper 2, Chemistry question 5