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For a reaction , `2N_(2)O_(5)rarr4NO_(2)+O_(2),` the rate is directly proportional to `[N_(2)O_(5)].` At `45^(@)C, 90%` of the `N_(2)O_(5)` react in 3600 s. The value of the rate constant isA. `3.2xx10^(-4)s^(-1)`B. `6.4xx10^(-4)s^(-1)`C. `8.5xx10^(-4)s^(-1)`D. `12.8xx10^(-4)s^(-1)` |
Answer» Correct Answer - B Reaction is first order. So, `In(a)/(a-x)=kt` Where, a=initial concentration =100 (left) a-x =concentration at time t=10 `(becausex=90%)` `In(100)/(10)=kxx3600" "(t=3600s)` `rArr2.303log10=kxx3600as(log10=1)` `k=(2.303)/(3600)=6.4xx10^(-4)s^(-1)` |
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