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The value of `K_(c)` for the following equilibrium is `CaCO_(3(s))hArrCaO_((s))+CO_(2(g))`. Given `K_(p)=167` bar at 1073 K.A. `1.896" mol L"^(-1)`B. `4.38xx10^(-4)" mol L"^(-1)`C. `6.3xx10^(-4)" mol L"^(-1)`D. `6.626" mol L"^(-1)` |
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Answer» Correct Answer - A `K_(p)=K_(c)(RT)^(Deltan)," "Deltan=1` `K_(p)=167" bar"`, `K_(c)=(167" bar")/(0.0821"L bar K"^(-1)"mol"^(-1)xx1073K)=1.896" mol L"^(-1)` |
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