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Standard free energy change of a reaction is +150 kJ mol^(-1). Calculate K_(p) at 30^(@)C. |
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Answer» Solution :`DELTA G^(@) = -2.303 RT log_(10) K_(p)` `log_(10) K_(p) = (Delta G^(@))/(-2.303 xx R xx T)` `log_(K_(p)) = (150000)/(-2.303 xx 8.314 xx 300)` `log K_(p) = -26.1136` `K_(p) = "ANTILOG"- 26.1136` `K_(p) = "antilog" BAR(27).8864` `= 7.698 xx 10^(-27)` |
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