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The equilibrium constant of a reaction is `1xx10^(20)` at `300 K`. The standard free energy change for this reaction is :A. `-115kJ`B. `+115 kJ`C. `+166 kJ`D. `-166 kJ` |
Answer» `DeltaG^(@)=-2.303RT log K_(p)` `=-2.303xx8.314xx300xxlog10^(20)J` `=-114.88 kJ` |
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