JEE Main 2019 — Chemistry Physical Chemistry

2019mcqhard

For the reaction $2A+B\rightarrow C$ , the values of initial rate at different reactant concentrations are given in the table below. The rate law for the reactions is:

$[A](mol{L}^{-1})$	$[B](mol{L}^{-1})$	$Initial Rate$ $(mol{L}^{-1}{s}^{-1})$
0.05	0.05	0.045
0.10	0.05	0.090
0.20	0.10	0.72

Official previous-year question

Held on 8 Apr 2019 · Verified 6 Jul 2026.

Options

A
$Rate=k{[A]}^{2}[B]$
B
$Rate=k{[A]}^{2}{[B]}^{2}$
C
$Rate=k[A][B]$
D
$Rate=k[A]{[B]}^{2}$

Did you get this right?

Solution

From the experiment (1) and (2)

${(\frac{0.05}{0.10})}^{x}{(\frac{0.05}{0.05})}^{y}=\frac{0.045}{0.090}$

${(\frac{1}{2})}^{x}=\frac{1}{2}\Rightarrow x=1$

From the experiment (2) and (3)

${(\frac{0.1}{0.2})}^{x}{(\frac{0.05}{0.1})}^{y}=(\frac{0.09}{0.72})$

$(\frac{1}{2}){(\frac{1}{2})}^{y}=\frac{0.09}{0.72}$

${(\frac{1}{2})}^{y}=\frac{0.09\times 2}{0.72}$

${(\frac{1}{2})}^{y}=(\frac{1}{4})$

$y=2$

Order of reaction with respect to A is $2$ and $B=2$

$\therefore R\propto [A]{[B]}^{2}$

Or $R=K[A]{[B]}^{2}$

Your note

JEE Main Physical Chemistry in other years

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Work through every JEE Main Physical Chemistry PYQ, year by year.

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JEE Main 2019 — Chemistry Physical Chemistry

2019mcqhard

For the reaction $2A+B\rightarrow C$ , the values of initial rate at different reactant concentrations are given in the table below. The rate law for the reactions is:

$[A](mol{L}^{-1})$	$[B](mol{L}^{-1})$	$Initial Rate$ $(mol{L}^{-1}{s}^{-1})$
0.05	0.05	0.045
0.10	0.05	0.090
0.20	0.10	0.72

Official previous-year question

Held on 8 Apr 2019 · Verified 6 Jul 2026.

Options

A
$Rate=k{[A]}^{2}[B]$
B
$Rate=k{[A]}^{2}{[B]}^{2}$
C
$Rate=k[A][B]$
D
$Rate=k[A]{[B]}^{2}$

Did you get this right?

Solution

From the experiment (1) and (2)

${(\frac{0.05}{0.10})}^{x}{(\frac{0.05}{0.05})}^{y}=\frac{0.045}{0.090}$

${(\frac{1}{2})}^{x}=\frac{1}{2}\Rightarrow x=1$

From the experiment (2) and (3)

${(\frac{0.1}{0.2})}^{x}{(\frac{0.05}{0.1})}^{y}=(\frac{0.09}{0.72})$

$(\frac{1}{2}){(\frac{1}{2})}^{y}=\frac{0.09}{0.72}$

${(\frac{1}{2})}^{y}=\frac{0.09\times 2}{0.72}$

${(\frac{1}{2})}^{y}=(\frac{1}{4})$

$y=2$

Order of reaction with respect to A is $2$ and $B=2$

$\therefore R\propto [A]{[B]}^{2}$

Or $R=K[A]{[B]}^{2}$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2019 — 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

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Keep practicing Physical Chemistry

JEE Main 2019 — 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