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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 : |
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Answer» `1.5xx10^(-4)` `-(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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