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)

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 negative at all temperature.
Hence, the reaction is thermodynamically favourable.
As the reaction is an exothermic process due to negative 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.