A) 0
B) 1
C) 2
D) 3
Correct Answer: C
Solution :
Let the order of the reaction is n. \[{{r}_{1}}=k{{[A]}^{n}}\] \[{{r}_{2}}=k{{[2A]}^{n}}\] \[\because \] \[\frac{{{r}_{2}}}{{{r}_{1}}}=\frac{{{[2A]}^{n}}}{{{[A]}^{n}}}\] \[4=\frac{{{[2A]}^{n}}}{{{[A]}^{n}}}\] \[{{2}^{2}}={{2}^{n}}\] \[n=2\]You need to login to perform this action.
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