JEE Main 2014 — Chemistry Physical Chemistry

2014mcqmedium

For the non-stoichiometry reaction, $2 \text{A} + \text{B} \rightarrow \text{C} + \text{D}$ , the following kinetic data were obtained in three separate experiments, all at $298K$ .

Initial Concentration (A)	Initial Concentration (B)	Initial rate of formation of C $( {\text{mol L}}^{-} {S}^{-} )$
$\text{0.1 M}$	$\text{0.1 M}$	$\text{1.2} \times 1 {0}^{ - 3 }$
$\text{0.1 M}$	$\text{0.2 M}$	$\text{1.2} \times 1 {0}^{ - 3 }$
$\text{0.2 M}$	$\text{0.1 M}$	$\text{2.4} \times 1 {0}^{ - 3 }$

The rate law for the formation of $C$ is

Official previous-year question

Held on 6 Apr 2014 · Verified 6 Jul 2026.

Options

A
$\frac{ \text{dc} }{ \text{dt} } = \text{k} [ \text{A} ] [ \text{B} ]$
B
$\frac{ \text{dc} }{ \text{dt} } = \text{k} { [ \text{A} ] }^{2} [ \text{B} ]$
C
$\frac{ \text{dc} }{ \text{dt} } = \text{k} [ \text{A} ] { [ \text{B} ] }^{2}$
D
$\frac{ \text{dc} }{ \text{dt} } = \text{k} [ \text{A} ]$

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Solution

Let $x$ and $y$ be the orders of the reaction with respect to the reactants $A$ and $B$ .

${\mathrm{Rate}}_{(1)}=1.2\times {10}^{-3}={(0.1)}^{x}{(0.1)}^{y}.......(I) {\mathrm{Rate}}_{(2)}=1.2\times {10}^{-3}={(0.1)}^{x}{(0.2)}^{y}.......(\mathrm{II}) {\mathrm{Rate}}_{(3)}=2.4\times {10}^{-3}={(0.2)}^{x}{(0.1)}^{y}.......(\mathrm{III})$

$\frac{\text{II}}{\text{I}}={2}^{\text{y}}=1={2}^{0}\therefore \text{y}=0$

$\frac{ \text{III} }{ \text{I} } \Rightarrow {2}^{1} = { ( 2 ) }^{\text{x}} \therefore \text{x} = 1$

$\text{Rate} = \text{k} \cdot { [ \text{A} ] }^{1}$

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Initial Concentration (A)

Initial Concentration (B)

Initial rate of formation of C

( {\text{mol L}}^{-} {S}^{-} )

\text{0.1 M}

\text{0.1 M}

\text{1.2} \times 1 {0}^{ - 3 }

\text{0.1 M}

\text{0.2 M}

\text{1.2} \times 1 {0}^{ - 3 }

\text{0.2 M}

\text{0.1 M}

\text{2.4} \times 1 {0}^{ - 3 }

Options

$\frac{ \text{dc} }{ \text{dt} } = \text{k} [ \text{A} ] [ \text{B} ]$

$\frac{ \text{dc} }{ \text{dt} } = \text{k} { [ \text{A} ] }^{2} [ \text{B} ]$

$\frac{ \text{dc} }{ \text{dt} } = \text{k} [ \text{A} ] { [ \text{B} ] }^{2}$

$\frac{ \text{dc} }{ \text{dt} } = \text{k} [ \text{A} ]$

Solution

Let $x$ and $y$ be the orders of the reaction with respect to the reactants $A$ and $B$ .

$\frac{\text{II}}{\text{I}}={2}^{\text{y}}=1={2}^{0}\therefore \text{y}=0$

$\frac{ \text{III} }{ \text{I} } \Rightarrow {2}^{1} = { ( 2 ) }^{\text{x}} \therefore \text{x} = 1$

$\text{Rate} = \text{k} \cdot { [ \text{A} ] }^{1}$

JEE Main 2014 — Chemistry Physical Chemistry

Options

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Solution

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JEE Main 2014 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry