JEE Advance - Chemistry (2021 - Paper 2 Online - No. 3)

For the following reaction,
$$2X + Y\buildrel k \over \longrightarrow P$$ the rate of reaction is $${{d[P]} \over {dt}} = k[X]$$. 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)

The rate constant, k, of the reaction is 13.86 $$\times$$ 10^$$-$$4 s^$$-$$1

Half-life of X is 50 s.

At 50 s, $$-$$$${{d[X]} \over {dt}}$$ = 13.86 $$\times$$ 10^$$-$$3 mol L^$$-$$1 s^$$-$$1.

At 100 s, $$-$$$${{d[Y]} \over {dt}}$$ = 3.46 $$\times$$ 10^$$-$$3 mol L^$$-$$1 s^$$-$$1.

Explanation

As the concentration of reactant becomes half at t = 50 s. So, half-time of reaction is 50 s.

Given,

$${{dP} \over {dt}} = k{[X]^1}$$

$$ \Rightarrow - {1 \over 2}{{dX} \over {dt}} = {{dP} \over {dt}} = k{[X]^1}$$

$$ - {{dX} \over {dt}} = 2k{[X]^1}$$

$${{ - dX} \over {[X]}} = 2k\,dt$$

$$ - \int_{{X_0}}^X {{{dX} \over X} = 2k\,\int_0^t {dt} } $$

$$ - [\ln X]_{{X_0}}^X = 2k[t]_0^t$$

$$ - \ln X + \ln {X_0} = 2kt$$

$$\ln {{{X_0}} \over X} = 2kt$$

At t_1/2, $$X = {{{X_0}} \over 2}$$

$$\ln {{2{X_0}} \over {{X_0}}} = 2k{t_{1/2}}$$

$$k = {{\ln 2} \over {2{t_{1/2}}}} = {{0.693} \over {100}} = 6.93 \times {10^{ - 3}}{s^{ - 1}}$$

At t = 50 s, $$ - {{dx} \over {dt}} = 2k{[X]^1};[X] = 1$$ mol

So, $${{ - dX} \over {dt}} = 2 \times 6.93 \times {10^{ - 3}} \times 1 = 13.86 \times {10^{ - 3}}$$ mol L^-1 s^-1

At t = 100 s, $$ - {1 \over 2}{{dX} \over {dt}} = {{ - dY} \over {dt}} \Rightarrow {{ - dY} \over {dt}} = k{[Y]^1};[Y] = {1 \over 2}$$

$$ - {{dY} \over {dt}} = 6.93 \times {10^{ - 3}} \times {1 \over 2} = 3.46 \times {10^{ - 3}}$$ mol L^-1 s^-1

So, options (b), (c) and (d) are correct.

JEE Advance - Chemistry (2021 - Paper 2 Online - No. 3)

Explanation

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