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A.\[\Delta H\] is +ve, \[\Delta S\] is –ve.

B.\[\Delta H\] is -ve, \[\Delta S\] is +ve.

C.\[\Delta H\] is +ve, \[\Delta S\] is +ve and \[\Delta H < T\Delta S\]

D. \[\Delta H\] is +ve and\[\;\Delta H\; > T{\text{ }}\Delta S\].

Answer
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According to Gibbs free energy equation:

\[\Delta G{\text{ }} = {\text{ }}\Delta H - T\Delta S\]

\[\Delta G\] = Gibbs free energy change

\[\Delta H\] = enthalpy change

\[T\] = temperature

\[\Delta S\] = entropy change

The conditions for a process to be spontaneous are:

\[\Delta G < 0\] or \[\Delta G\] should always be negative.

For an endothermic reaction \[\Delta H{\text{ }} > {\text{ }}0\]or \[\Delta H\] is positive.

So for \[\Delta G\] to be negative, \[\Delta S\]must be positive and the \[T{\text{ }}\Delta S\] must be greater than \[\Delta H\] as the latter is positive.

Now considering the dissolution of \[N{H_4}Cl\], it is an endothermic reaction. Thus \[\Delta H{\text{ }} > {\text{ }}0\]or \[\Delta H\] is positive.

So \[\Delta S\] must be positive and the value of \[T{\text{ }}\Delta S\]must be greater than\[\Delta H\]. The dissolution of \[N{H_4}Cl\] results in a favorable increase of entropy which overcomes the enthalpy change during the dissolution. As a result \[T{\text{ }}\Delta S\] become negative and the total of \[\Delta H - T\Delta S\] is negative. Hence \[\Delta G\]is negative. This accounts for the favorable dissolution of ammonium chloride in water.

Unlike endothermic reaction, an exothermic reaction \[\Delta H\]is always negative and it makes the reaction go spontaneous. The change in entropy does not affect the spontaneity of the exothermic reaction.