QJEEM14Ch96dc: JEE Main 2014 Chemistry - Class 12 JEEM14Ch96dc Solution: For the non – stoichiometre reaction 2A + B C + D, the…

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For the non – stoichiometre reaction 2A + B

\to

C + D, the following kinetic data were obtained in three separate experiments, all at 298 K.

Initial Concentration (A)	Initial Concentration (B)	Initial rate of formation of C (mol L^-1 s^-1)
0.1 M	0.1 M	1.2 x 10^-3
0.1 M	0.2 M	1.2 x 10^-3
0.2 M	0.1 M	2.4 x 10^-3

The rate law for the formation of C is:

Accepted Answer

Let rate of reaction

= {{d\left[ C \right]} \over t} = k{\left[ A \right]^x}{\left[ B \right]^y}

Now from the given data

1.2 \times {10^{ - 3}} = k{\left[ {0.1} \right]^x}{\left[ {0.1} \right]^y}\,\,\,\,\,...\left( i \right)

1.2 \times {10^{ - 3}} = k{\left[ {0.1} \right]^x}{\left[ {0.2} \right]^y}\,\,\,\,\,...\left( {ii} \right)

2.4 \times {10^{ - 3}} = k{\left[ {0.2} \right]^x}{\left[ {0.1} \right]^y}\,\,\,\,...\left( {iii} \right)

Dividing equation

(i)

by

(ii)

\Rightarrow \,\,\,\,{{1.2 \times {{10}^{ - 3}}} \over {1.2 \times {{10}^{ - 3}}}} = {{k{{\left[ {0.1} \right]}^x}{{\left[ {0.1} \right]}^y}} \over {k{{\left[ {0.1} \right]}^x}{{\left[ {0.2} \right]}^y}}}

We find,

y=0

Now dividing equation

(i)

by

(iii)

\Rightarrow \,\,\,\,{{1.2 \times {{10}^{ - 3}}} \over {2.4 \times {{10}^{ - 3}}}} = {{k{{\left[ {0.1} \right]}^x}{{\left[ {0.1} \right]}^y}} \over {k{{\left[ {0.2} \right]}^x}{{\left[ {0.1} \right]}^y}}}

We find,

x=1

Hence

{{d\left[ C \right]} \over {dt}} = k{\left[ A \right]^1}{\left[ B \right]^0}

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