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InterviewSolution
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The standard enthalpy and entropy changes for the reaction in equilibrium for the forward direction are given below: `CO_((g))+H_(2)O_((g))hArrCO_(2(g))+H_(2(g))` `DeltaH_(300K)^(@)=-41.16 kJ mol^(-1)` `DeltaS_(300 K)^(@)=-4.24xx10^(-2) kJ mol^(-1)` `DeltaH_(1200 K)^(@)=-32.93 K J mol^(-1)` `DeltaS_(1200 K)^(@)=-2.96xx10^(-2) k J mol^(-1)` Calculate `K_(p)` at each temperature and predict the direction of reaction at `300 K` and `1200 K`, when `P_(CO)=P_(CO_(2))=P_(H_(2))=P_(H_(2)O)=1` atm at initial state. |
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Answer» Correct Answer - At `300 K,DeltaG^(@)=-28.44 kJ,K_(p)=8.94xx10^(4)` `at 1200 K,DeltaG^(@)=+2.59 kJ,K_(p)=0.77` |
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