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

For the reaction:
`CO(g) +H_(2)O(g) hArr CO_(2)(g) +H_(2)(g)`
`(Delta_(r)H)_(300K) = - 41.2 kJ mol^(-1)`
`(Delta_(r)H)_(1200K) =- 33.0 kJ mol^(-1)`
`(Delta_(r)S)_(300K) = - 4.2 xx 10^(-2) kJ mol^(-1)`
`(Delta_(r)S)_(1200K) =- 3.0 xx10^(-2) kJ mol^(-1)`
Predict the direction of spontaneity of the reaction at `300K` and `1200K`. also calculated `log_(10)K_(p)` at `300K` and `1200K`.

Text Solution

Verified by Experts

Using : `Delta_(r )G=Delta_(r )H-T Delta_(r )S`
At 1200 K : `(Delta_(r )G)_(300 K)=-41.2 -300 xx(-4.2xx10^(-2))=-28.6 kJ mol^(-1)`
At 1200 K : `(DeltA_(r )G)_(1200 K)=-33.0 -1200xx(-3.0xx10^(-2))=3 kJ mol^(-1)`
Since, `Delta_(r )G` is negative at 300 K and positive at 1200 K, reaction is spontneous at 300 K and non - spontaneous at 1200 K.
Also, `Delta_(r )G^(Θ)=-RT ln K_(eq)`
Since, `Delta_(r ) G^(Θ)` is not given, assume `(Delta_(r )G^(Θ))~~(Delta_(r )G)_(300 K)`
At 300 K `-28.6=-(8.314xx10^(-3))xx300 ln K_(eq) " " rArr " " log_(10)(K_(eq))_(300 K)=4.98 [because ln x = 2.303 log_(10)x]`
and at 1200 K : `-28.6 = -(8.134xx10^(-3))xx1200 ln K_(eq)" "rArr " " log_(10)(K_(eq))_(1200 K)=1.24`
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