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) \[{{\log }_{10}}{{[{{N}_{2}}{{O}_{5}}]}_{t}}={{\log }_{10}}{{[{{N}_{2}}{{O}_{5}}]}_{0}}-kt\]
D) \[\text{In}\frac{{{\text{ }\!\![\!\!\text{ }{{\text{N}}_{\text{2}}}{{O}_{5}}]}_{0}}}{{{\text{ }\!\![\!\!\text{ }{{\text{N}}_{\text{2}}}{{O}_{5}}]}_{t}}}=kt\]
Correct Answer: D
Solution :
Rate constant \[=2.3\times {{10}^{-2}}{{\sec }^{-1}}\] It means it is a first order reaction (because unit of rate constant is sec?1) For first order reaction \[K=\frac{1}{t}\ln \frac{a}{a-x}\] \[Kt=\ln \frac{a}{a-x}=\ln \frac{{{[{{N}_{2}}{{O}_{5}}]}_{0}}}{{{[{{N}_{2}}{{O}_{5}}]}_{t}}}\]
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