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For the reaction `2NO_(2)rarrN_(2)O_(3)+O_(2)`, rate expression is as follows `:` `-(d[NO_(2)])/(dt)=K[NO_(2)]^(n)`, where `K=3xx10^(-3)mol^(-1)L sec^(-1)` If rate of formation of oxygen is `1.5xx10^(-4)mol L^(-1)sec^(-1)` then the molar concentration of `NO_(2)`in mole `L^(-1)` is `:`A. `1.5xx10^(-4)`B. `0.0151`C. `0.214`D. `0.316` |
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Answer» Correct Answer - 4 From the unit of k, it is evident that it is second order reaction. `-(1)/(2)=(d[NO_(2)])/(dt)=(d[O_(2)])/(dt)` `:." "-(d[NO_(2)])/(dt)=2xx(d[O_(2)])/(dt)=2xx1.5xx10^(-4)=3xx10^(-4)` `3xx10^(-4)=K[NO_(2)]^(2)=3xx10^(-3)[NO_(2)]^(2)` `:. " "[NO_(2)]=0.316` |
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