A) \[[{{N}_{2}}{{O}_{5}}]t={{[{{N}_{2}}{{O}_{5}}]}_{0}}+kt\]
B) \[{{[{{N}_{2}}{{O}_{5}}]}_{0}}={{[{{N}_{2}}{{O}_{5}}]}_{t}}{{e}^{kt}}\]
C) \[log{{[{{N}_{2}}{{O}_{5}}]}_{t}}=\log {{[{{N}_{2}}{{O}_{5}}]}_{0}}+kt\]
D) In \[\frac{{{[{{N}_{2}}{{P}_{5}}]}_{0}}}{{{[{{N}_{2}}{{P}_{5}}]}_{t}}}=kt\]
Correct Answer: D
Solution :
[d] As the unit of rate constant is \[{{\sec }^{-1}},\] so the reaction is first order reaction. Hence \[k=\frac{1}{t}\,\,ln\frac{a}{(a-x)}\]or \[kt=ln\frac{{{[{{N}_{2}}{{O}_{5}}]}_{0}}}{{{[{{N}_{2}}{{O}_{5}}]}_{t}}}\]You need to login to perform this action.
You will be redirected in
3 sec