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For the equilibrium: CO((g))+H(2)O((g)...

For the equilibrium:
`CO_((g))+H_(2)O_((g))hArrCO_(2(g))+H_(2(g))` the standard enthalpy and entropy changes at `300 K` and `1200 K` for the forward reaction are as follows:
`{:(DeltaH_(300 K)^(@)=-41.16kJmol^(-1),), (DeltaS_(300 K)^(@)=-0.0424kJmol^(-1)), (DeltaH_(1200 K)^(@)=-32.93kJmol^(-1),), (DeltaS_(1200K)^(@)=-0.0296kJmol^(-1)):}`
In which direction will the reaction be spontaneous?
(`a`) At `300 K`,
(`b`) At `1200 K`, when at equilibrium `P_(CO)=P_(CO_(2))=P_(H_(2))=P_(H_(2)O)=1atm`
Also calculate `K_(p)` for the reaction at each temperature.

Text Solution

Verified by Experts

(`a`) `K_(p)=8.94xx10^(4)` at `300K`,
(`b`) `K_(p)=0.773` at `1200K` ,
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