Question

Given: For ethylene oxide $\Delta H_f^{\circ }$ and $\Delta S^{\circ }$ are $-51{\text{kJ mol}}^{-1}$ and $243{\text{J mol}}^{-1}{\text{K}}^{-1}$ respectively and acetaldehyde, $\Delta H_f^{\circ }$ and $\Delta S^{\circ }$ are $-166{\text{kJ mol}}^{-1}$ and $266{\text{J mol}}^{-1}{\text{K}}^{-1}$ respectively, then select the correct statements :

• A. $\Delta H^{\circ }$ per the reaction is $-115{\text{kJ mol}}^{-1}$
• B. $\Delta S^{\circ }=0.023{\text{kJ mol}}^{-1}{\text{K}}^{-1}$
• C. Reaction is thermodynamically favourable
• D. Low temperature can favour formation of more product

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\Delta H^{\circ }$ per the reaction is $-115{\text{kJ mol}}^{-1}$
B $\Delta S^{\circ }=0.023{\text{kJ mol}}^{-1}{\text{K}}^{-1}$
C Reaction is thermodynamically favourable
D Low temperature can favour formation of more product

$\Delta H^{\circ }=\Delta H_f^{\circ }\text{(products)}-\Delta H_f^{\circ }\text{(reactants)}$
$=-166-(-51)=-115{\text{kJ mol}}^{-1}$
$\Delta S^{\circ }=266-243=23=0.023{\text{kJ mol}}^{-1}{\text{K}}^{-1}$
$\therefore \Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$
As $\Delta H^{\circ }is-ve$ and $\Delta S^{\circ }is+ve$ , then $\Delta G^{\circ }$ will be -ve at all temperature.
Hence, the reaction is thermodynamically favourable.
As the reaction is an exothermic process due to -ve value of enthalpy of reaction, then temperature on product side is more than the temperature on reactant side.
Since the reaction is exothermic , low temperature is favored.
So, more product will be formed.