JEE MAIN - Chemistry (2014 (Offline) - No. 22)
For the non – stoichiometre reaction 2A + B $$\to$$ C + D, the following kinetic data were obtained in three
separate experiments, all at 298 K.
The rate law for the formation of C is:
| 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 |
$${{dc} \over {dt}} = k[A]{[B]^2}$$
$${{dc} \over {dt}} = k[A]$$
$${{dc} \over {dt}} = k[A]{[B]}$$
$${{dc} \over {dt}} = k[A^2]{[B]}$$
Explanation
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 $${{d\left[ C \right]} \over {dt}} = k{\left[ A \right]^1}{\left[ B \right]^0}$$
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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