A) will diminish to 1/4 of initial value
B) will diminish to 1/8 of initial value
C) will increase 4 times
D) will increase 8 times
Correct Answer: B
Solution :
Rate \[\propto \] concentration \[\propto \frac{moles}{litre}\] When the volume of the reaction vessel is doubled then the rate of reaction \[({{r}^{1}})\] will be \[r=k\,{{\left[ \frac{x}{V} \right]}^{2}}\left[ \frac{y}{V} \right]\](where, x and y are the number of moles of NO and\[{{O}_{2}}\]) \[I=k{{\left[ \frac{x}{2V} \right]}^{2}}\left[ \frac{y}{2V} \right]\] \[\frac{I}{r}=\frac{k{{\left[ \frac{x}{2V} \right]}^{2}}\left[ \frac{y}{2V} \right]}{k{{\left[ \frac{x}{V} \right]}^{2}}\left[ \frac{y}{V} \right]}\Rightarrow r\frac{1}{8}\,r\]
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