Question

For the reaction, $X_2{O}_4\left(l\right)⟶2X{O}_2\left(g\right)$
$\Delta U=2.1kcal$, $\Delta S=20cal{K}^{-1}$ at 300 K Hence , $\Delta G$ is_________.

• A. +2.7 kcal
• B. -2.7 kcal
• C. +9.3 kcal
• D. -9.3 kcal

Answer 1

Answer: (B)

-2.7 kcal

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 reaction

Thus in order to determine $\Delta G$, the value of $\Delta H$ must be known. The value of $\Delta H$ can be calculated by the reaction,$\Delta H=\Delta U+\Delta n_{g}RT$

where,

$\Delta U=$change in internal energy

$\Delta n_g=$(number of moles of gaseous product)-(number of moles of gaseous reactant)$=2-0=2$

$R=$gas constant$=2cal$

$\Delta U=2.1kcal=2.1\times {10}^{3}cal$

therefore,

$\Delta H=(2.1\times {10}^3)+(2\times 2\times 300)=3300cal$

Hence,

$\Delta G=\Delta -T\Delta S$

$\Delta G=3300-(300\times 20)$

$\Delta G=-2700cal$

$\Delta G=-2.7kcal$