For the following reaction \[ 2 \mathbf{X}+\mathbf{Y} \xrightarrow{k} \mathbf{P} \] the rate of reaction is $\frac{\mathrm{d}[\mathbf{P}]}{\mathrm{dt}}=k[\mathbf{X}]$. 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: $\ln 2=0.693$ )

(B) Half-life of $\mathbf{X}$ is 50 s .; (C) At $50 \mathrm{~s},-\frac{\mathrm{d}[\mathbf{X}]}{\mathrm{dt}}=13.86 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$.; (D) At $100 \mathrm{~s},-\frac{\mathrm{d}[\mathbf{Y}]}{\mathrm{dt}}=3.46 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$.

(A) The rate constant, $k$, of the reaction is $13.86 \times 10^{-4} \mathrm{~s}^{-1}$.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Chemical Kinetics

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For the following reaction

2 X + Y k P

the rate of reaction is

\frac{d [ P ]}{dt} = k [X]

. Two moles of

X

are mixed with one mole of

Y

to make 1.0 L of solution. At 50 s, 0.5 mole of

Y

is left in the reaction mixture. The correct statement(s) about the reaction is(are)
(Use:

ln 2 = 0.693

)

(A)The rate constant, $k$ , of the reaction is $13.86 \times 1 0^{- 4} s^{- 1}$ .
(B)Half-life of $X$ is 50 s .✓ Correct
(C)At $50 s, - \frac{d [ X ]}{dt} = 13.86 \times 1 0^{- 3} mol L^{- 1} s^{- 1}$ .✓ Correct
(D)At $100 s, - \frac{d [ Y ]}{dt} = 3.46 \times 1 0^{- 3} mol L^{- 1} s^{- 1}$ .✓ Correct

Answer: (B), (C), (D)

Solution

The reaction is first-order with respect to X. From stoichiometry, when 0.5 mol of Y reacts, 1 mol of X reacts, leaving [X] = 1 M at t = 50 s. For a first-order reaction,

- \frac{1}{2} \frac{d [ X ]}{d t} = k [X]

, so

- \frac{d [ X ]}{d t} = 2 k [X]

. Since [X] goes from 2 M to 1 M in 50 s, the half-life is 50 s (option B correct). From

t_{1/2} = \frac{l n 2}{2 k}

, we get

2 k = \frac{0.693}{50}

, so

k = 6.93 \times 1 0^{- 3}

^{- 1}

. At 50 s,

- \frac{d [ X ]}{d t} = 2 k \times 1 = 2 \times 6.93 \times 1 0^{- 3} = 13.86 \times 1 0^{- 3}

mol L

^{- 1}

^{- 1}

(option C correct). At 100 s, [X] = 0.5 M, so

- \frac{d [ Y ]}{d t} = k [X] = 6.93 \times 1 0^{- 3} \times 0.5 = 3.465 \times 1 0^{- 3}

mol L

^{- 1}

^{- 1}

(option D correct). Option A is incorrect as it states

k = 13.86 \times 1 0^{- 4}

instead of

6.93 \times 1 0^{- 3}

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JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Chemical Kinetics

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