Question

For the reaction \[{{\rm{A}}_{{\rm{(l)}}}} \to {\rm{2}}{{\rm{B}}_{\left( {\rm{g}} \right)}}\]\[\Delta {\rm{U}}\]= 2.1 kcal, \[\Delta {\rm{S}}\]= 20 cal/K at 300 K. Hence Gibbs free energy change in kcal is

Answer 1

Hint: In this question, we are given the entropy change, internal energy change and temperature. To find the Gibbs free energy change, we have to first calculate the enthalpy change.

Formula used \[\Delta {\rm{H = }}\Delta {\rm{U-}}\Delta {{\rm{n}}_{{\rm{(g)}}}}{\rm{RT}}\]
\[\Delta {\rm{G = }}\Delta {\rm{H-T}}\Delta {\rm{S}}\]
where
\[\Delta {\rm{H}}\]= enthalpy change
\[\Delta {\rm{G}}\]=Gibbs free energy change
\[\Delta {\rm{S}}\]=entropy change
T=temperature
\[\Delta {{\rm{n}}_{{\rm{(g)}}}}\]=the difference in the no.of moles of products and reactants in the gaseous state.

Complete Step by Step Solution:
In the given reaction,
\[\Delta {{\rm{n}}_{{\rm{(g)}}}}\]=no.of moles of product in a gaseous state- no.of moles of reactant in a gaseous state.
The reactant is in a liquid state.
So, \[\Delta {{\rm{n}}_{{\rm{(g)}}}} = 2 - 0 = 2\]
We have to find the enthalpy change now.
\[\Delta {\rm{U = 2}}{\rm{.1kcal = 2100cal}}\]
T=300K
R=1.98Cal/mol/K
\[\Delta {\rm{H = }}\Delta {\rm{U-}}\Delta {{\rm{n}}_{{\rm{(g)}}}}{\rm{RT}}\]
\[{\rm{ = 2100cal + 2moles}}\left( {{\rm{1}}{\rm{.98cal/mol/K}}} \right)\left( {{\rm{300K}}} \right)\]
=3288 cal

Now, we have the value of enthalpy change, entropy change, and temperature, so we can calculate Gibbs's free energy change.


T=300K
So,
\[\Delta {\rm{G = }}\Delta {\rm{H-T}}\Delta {\rm{S}}\]
\[{\rm{ = 3288cal - (300K}} \times 20{\rm{cal/K)}}\]
= -2712 cal

So, the Gibbs free energy change is -2712 cal.

Additional information: Gibbs free energy change is the net energy accessible to do valuable work and is, therefore, an estimate of the ‘free energy. Due to this, it is also recognized as the free energy of the reaction.
Gibbs's free energy change provides a measure of spontaneity at constant pressure and temperature.
(i) If Gibbs's free energy change is negative, the process is spontaneous.
(ii) If Gibbs's free energy change is positive, the process is nonspontaneous.

Note: While attempting the question, one must pay attention to the equation in which the reactant is in a liquid state and the product in a gaseous state. So, only the moles of the product will be calculated while calculating the no.of moles. Units of the given values and the final answer must be mentioned in the answer.