A) 2
B) 0
C) 1
D) 1.5
Correct Answer: C
Solution :
Given, rate \[=k{{[{{H}^{+}}]}^{n}}\] Initial \[pH=3\]so \[[{{H}^{+}}]=1\times {{10}^{-3}},\]initial rate \[={{r}_{1}}\] Final pH = 1 so \[[{{H}^{+}}]=1\times {{10}^{-1}},\] Final rate \[={{r}_{2}}=100{{r}_{1}}\] On substituting values, we get \[{{r}_{1}}=k{{[1\,\times \,{{10}^{-3}}]}^{n}}\] ?(i) \[{{r}_{2}}=100\,{{r}_{1}}=k{{[{{10}^{-1}}]}^{n}}\] ?(ii) Dividing eqs (i) by (ii), \[\frac{{{r}_{1}}}{100{{r}_{1}}}={{\left[ \frac{1\times {{10}^{-3}}}{1\times {{10}^{-1}}} \right]}^{n}}\] \[\frac{1}{100}={{\left[ \frac{{{10}^{-2}}}{1} \right]}^{n}}\Rightarrow {{\left[ \frac{1}{100} \right]}^{1}}={{\left[ \frac{1}{100} \right]}^{n}}\] \[\therefore \] \[n=1\] Thus, the reaction is of first order.
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