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For the reaction, `X_(2)O_(4)(l)rarr2XO_(2)(g),DeltaE=2.1Kcal` , `DeltaS=20cal//K` at `300K` . Hence `DeltaG` isA. `2.7Kcal`B. `-2.7Kcal`C. `9.3Kcal`D. `-9.3Kcal` |
Answer» Correct Answer - B From `DeltaH=DeltaE+DeltanRT` `DeltaH=2.1+2xx(2xx10^(-3))xx300` `3.3` Now from `DeltaG=DeltaH-TDeltaS` `=3.3-300xx(20xx10^(-3))` `=-2.7Kcal` |
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