A) \[\Delta S\] is negative and therefore, \[\Delta H\] should be highly positive
B) \[\Delta S\] is negative and therefore, \[\Delta H\] should be highly negative
C) \[\Delta S\] is positive and therefore, \[\Delta H\] should be negative
D) \[\Delta S\] is positive and therefore, \[\Delta H\] should also be highly positive.
Correct Answer: B
Solution :
| \[\Delta S\][change in entropy] and \[\Delta H\] [change in enthalpy] are related by the equation |
| \[\Delta G=\Delta H-T\Delta S\] |
| [here, \[\Delta G\] = change in Gibbs free energy] |
| For adsorption of a gas, \[\Delta S\] is negative because randomness decreases. Thus, in order to make \[\Delta G\] negative [for spontaneous reaction], \[\Delta H\] must be highly negative. Hence, for the adsorption of a gas, if ]\[\Delta S\] is negative, therefore, \[\Delta H\] should be highly negative. |
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