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Calculate the standard Gibb's energy change for a reaction at 298 K, if its equilibrium constant is 50. |
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Answer» Solution :`Delta G^(@) = -2.303 T log K` `= -2.303 XX 8.314 xx 298 xx log 50` `= -2.303 xx 8.314 xx 298 xx 1.6990` `= -9694 J = -9.694 KJ mol^(-1)`. |
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