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The equilibrium constant of a reaction is `0.008` at `298 K`. The standard free energy change of the reaction at the same temperture isA. `+11.96 kJ`B. `-11.96 kJ`C. `-5.43 kJ`D. `-8.46 kJ` |
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Answer» Correct Answer - A `DeltaG^(@)=-2.303 RT log K` `=-2.303xx8.314xx298xxlog0.008` `=-2.303xx8.314xx298xx(-2.0969)` `=11964.6 J=11.96 kJ`. |
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