JEE Advanced 2021 · Paper 2 · Chemistry Q3

← JEE Advanced 2021 · Paper 2

ChemistryMultiple correct+4 / -2Question 22 of 57

For the following reaction
$2\mathbf X + \mathbf Y \xrightarrow{k} \mathbf P$
the rate of reaction is $\dfrac{d[\mathbf P]}{dt} = k[\mathbf X]$ . Two moles of $\mathbf X$ are mixed with one mole of $\mathbf Y$ to make 1.0 L of solution. At 50 s, 0.5 mole of $\mathbf Y$ is left in the reaction mixture. The correct statement(s) about the reaction is(are)

(Use: $\ln 2 = 0.693$ )

The rate constant,

k

, of the reaction is

13.86 \times 10^{-4}

^{-1}

Half-life of

\mathbf X

is 50 s.

At 50 s,

-\dfrac{d[\mathbf X]}{dt} = 13.86 \times 10^{-3}

mol L

^{-1}

^{-1}

At 100 s,

-\dfrac{d[\mathbf Y]}{dt} = 3.46 \times 10^{-3}

mol L

^{-1}

^{-1}

Answer and worked solution

Official answer: B, C, D

Rate law for X. $-\dfrac{d[\mathbf X]}{dt}=2\dfrac{d[\mathbf P]}{dt}=2k[\mathbf X]$ , so $[\mathbf X]=2e^{-2kt}$ .

Find k. At 50 s, 0.5 mol of Y has reacted, so 1 mol of X has reacted and $[\mathbf X]=1$ M. Then $e^{-100k}=\frac12$ , so
$k=6.93\times10^{-3}\text{ s}^{-1}.$

Check each option.

(A) $13.86\times10^{-4}$ s $^{-1}$ is not $k$ . False.
(B) The half-life of X is $\dfrac{\ln2}{2k}=50$ s. True.
(C) At 50 s, $-\dfrac{d[\mathbf X]}{dt}=2k(1)=13.86\times10^{-3}$ mol L $^{-1}$ s $^{-1}$ . True.
(D) At 100 s, $[\mathbf X]=0.5$ M, so $-\dfrac{d[\mathbf Y]}{dt}=k[\mathbf X]=3.46\times10^{-3}$ mol L $^{-1}$ s $^{-1}$ . True.

(B), (C), (D)

Topics:Integrated rate equations Half-life Rate of reaction and rate law

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JEE Advanced 2021 · Paper 2, Chemistry question 3