A) 0
B) 1
C) 2
D) 3
Correct Answer: C
Solution :
\[r=k{{[{{H}^{+}}]}^{n}}\] Exp.(l)\[{{r}_{1}}=k\,\,{{[{{10}^{-2}}]}^{n}}\] (If \[pH=2\] then\[[{{H}^{+}}]={{10}^{-2}}\]) Exp. (II) \[100{{r}_{1}}=k\,\,{{[{{10}^{-1}}]}^{n}}\](If \[pH=1\] then \[[{{H}^{+}}]={{10}^{-1}}\]) \[\frac{Exp.(II)}{Exp.(II)}=\frac{100{{r}_{1}}}{{{r}_{1}}}=\frac{k{{[0.1]}^{n}}}{k{{[0.01]}^{n}}}=\frac{100}{1}={{\left[ \frac{10}{1} \right]}^{n}}\]or\[n=2\]Order of reaction is 2.You need to login to perform this action.
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