Answer:

Correct Answer: B

Solution:

The change in Gibbs free energy is given by $ \Delta G=\Delta H-T\Delta S $ where, $ \Delta H $ = enthalpy of the reaction $ \Delta S $ = entropy of the reaction

Thus, in order to determine $ \Delta G $ , the values of $ \Delta H $ must be known, the value of $ \Delta H $ can be calculated by the equation $ \Delta H=\Delta U+\Delta n _{g}RT $

where $ \Delta U $ = change in internal energy $ \Delta n _{g} $ = (number of moles of gaseous products) - (number of moles of gaseous reactants) = 2 - 0 = 2 R = gas constant = 2 cal But,

$ \Delta H=\Delta u+\Delta n _{g}RT $ $ \Delta u=2.1kcal=2.1\times 10^{3}cal $ $ [\because ,1kcal,=10^{3}cal] $

$ \therefore $ $ \Delta H=(2.1\times 10^{3})+(2\times 2\times 300)=3300cal $ Hence, $ \Delta G=\Delta H-T\Delta S $

Þ $ \Delta G=(3300)-(300-20) $ $ \Delta G=-2700,cal $

$ \therefore $ $ \Delta G=-2.7kcal $