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The equilibrium constant for, `2H_(2)S(g)hArr2H_(2)(g)+S_(2)(g) "is" 0.0118 "at" 1300 K` while the heat of dissociation is `597.4KJ`. The standard equilibrium constant of the reaction at `1200K` is:A. `1.180xx10^(-4)`B. `11.80`C. `118.0`D. cannot be calculated from given data |
Answer» Correct Answer - A `log(K_(2))/(K_(1))=(DeltaH^(@))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]` `log(K_(2))/(0.0118)=(597.4xx10^(3))/(2.303xx8.314)[(1)/(1300)-(1)/(1200)]` `log K_(2)=-2+log(0.0118)=-3.928 rArr K_(2)=1.18xx10^(-4)` Therefore. (A) option is correct. |
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