Question

For the reaction :
$\ce{X_2 O_4(l) -> 2XO_2(g) }$
$\ce{\Delta U = 2.1 \, k \, cal , \Delta S = 20 \, cal \, k^{-1} }$ at 300 K
Hence $\Delta G$ is : -

• A. 2.7 k cal
• B. - 2.7 k cal
• C. 9.3 k cal
• D. - 9.3 k cal

Answer 1

Answer: (B)

The change in Gibbs Free energy is given by $\Delta H = \Delta U + \Delta n _{ g } R T$
where $\Delta H$ is the enthalpy of the reaction
$\Delta S$ is the entropy of the reaction and $\Delta U$ is the change in internal energy $\Delta n _{ g }$ is the (number of gaseous moles in product) - (number of gaseous moles in reactant )$=2-0=2$
$R$ is the gas constant $=2$ cal But, $\Delta H =\left(2.1 \times 10^{3}\right)+(2 \times 2 \times 300)=3300 cal$
Hence, $\Delta G =\Delta H = T \Delta S$
$\Delta G=3300-(300 \times 20)$
$\Delta G =- 2 7 0 0 c a l =- 2.7 c a l$