A) 1.5
B) 2.0
C) 2.5
D) 3.0
Correct Answer: B
Solution :
\[aA\xrightarrow{{}}xP\] - rate \[\propto {{[A]}^{a}}=k{{[A]}^{a}}\] Initially \[{{[A]}_{1}}=2.2\] millimol, \[{{r}_{1}}=2.4mM/s\], If \[{{[A]}_{2}}=\frac{2.2}{2}=1.1\] millimol, \[{{r}_{2}}=0.6mM/s\] We have \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{[A]_{1}^{a}}{{{[{{A}_{2}}]}^{a}}}\] (a is the order). \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{{{(2.2)}^{a}}}{{{(1.1)}^{a}}}\] or \[\frac{2.4}{0.6}={{(2)}^{a}}\] \[4={{2}^{a}}\] \[\Rightarrow \] \[{{2}^{2}}={{2}^{a}}\] or \[a=2\] The reaction is of second order.You need to login to perform this action.
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