In the following reaction; $x A \longrightarrow y B$ $\log_{10}\left[-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\right]=\log_{10}\left[\frac{\mathrm{~d}[\mathrm{~B}]}{\mathrm{dt}}\right]+0.3010$ 'A' and 'B' respectively can be :

(D) $\mathrm{C}_{2} \mathrm{H}_{4}$ and $\mathrm{C}_{4} \mathrm{H}_{8}$

(A) $\mathrm{C}_{2} \mathrm{H}_{2}$ and $\mathrm{C}_{6} \mathrm{H}_{6}$

(C) $\mathrm{N}_{2} \mathrm{O}_{4}$ and $\mathrm{NO}_{2}$

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Chemical Kinetics

In the following reaction;

x A ⟶ y B

lo g_{10} [- \frac{d [ A ]}{dt}] = lo g_{10} [\frac{d [ B ]}{dt}] + 0.3010

'A' and 'B' respectively can be :

Answer: (D)

For the reaction

x A ⟶ y B

the rate is one and the same when referred to either species, so

- \frac{1}{x} \frac{d [ A ]}{d t} = \frac{1}{y} \frac{d [ B ]}{d t}

, which rearranges to

- \frac{d [ A ]}{d t} = \frac{x}{y} \frac{d [ B ]}{d t}

. Taking logarithms,

lo g_{10} [- \frac{d [ A ]}{d t}] = lo g_{10} [\frac{d [ B ]}{d t}] + lo g_{10} \frac{x}{y}

. Comparing with the given relation,

lo g_{10} \frac{x}{y} = 0.3010 = lo g_{10} 2

, so

\frac{x}{y} = 2

: the reactant's coefficient must be twice the product's. The dimerisation

2 C X_{2} H X_{4} C X_{4} H X_{8}

has

x = 2

and

y = 1

, giving

\frac{x}{y} = 2

exactly, whereas

N X_{2} O X_{4} 2 NO X_{2}

has

x = 1, y = 2

and so gives the reciprocal

\frac{1}{2}

. The pair that fits is ethene and its dimer,

C X_{2} H X_{4}

and

C X_{4} H X_{8}

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option