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Delta G^(@) of a reaction is 120 kJ mol^(-1). Calculate the K_(p) at 20^(@)C. |
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Answer» Solution :`Delta G^(@) = -2.303 RT log_(10) K_(p)` `-120,000 = -2.303 xx 8.314 xx 293 xx log_(10)K_(p)` `log_(10)K_(p) = (-120,000)/(-2.303 xx 8.314 xx 293)` `log_(10)K_(p) = 21.39` `K_(p) = "ANTILOG" 21.39` `- 2.455 xx 10^(21)` |
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