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For the reaction equlibrium `N_(2)O_(4)hArr2NO_(2(g))`the concentration of `N_(2)O_(4) and NO_(2)` at equlibrium are `4.8xx10^(-2)and 1.2xx10^(-2)"mol lite"^(-1)` respectively. The value of `K_(c)` for the reaction isA. `3.3xx10^(2)"mol litre"^(-1)`B. `3xx10^(-1)"mol litre"^(-1)`C. `3xx10^(-3)"mol litre"^(-1)`D. `3xx10^(3)"mol litre"^(-1)` |
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Answer» Correct Answer - C `K=([NO_(2)]^(2))/([N_(2)O_(4)])=([1.2xx10^(-2)])/([4.8xx10^(-2)])=0.3xx10^(-2)=3xx10^(-3)` |
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