Question

The enthalpy change for the conversion of \(\frac{1}{2} Cl_2(g)\) to \(Cl^{-}(aq)\) is (-)_____________ \(KJ mol^{-1}\) (Nearest integer)
Given : \(\Delta_{\text {dis }} H _{ Cl _{2( g )}}^{\ominus}=240\, kJ \,mol ^{-1}\), \(\Delta_{ eg } H _{ Cl }^{\ominus}=-350\, kJ \,mol ^{-1}\), \(\Delta_{\text {hyd }} H ^\theta Cl _{( g )}^{-}=-380 \,kJ \,mol ^{-1}\)

Hint:

The enthalpy change for a reaction can be determined by using the Hess's Law, which states that the total enthalpy change is the sum of enthalpy changes of individual reactions.

Answer 1

Correct Answer: -610

\(\frac{1}{2}​Cl_{2(g)}​→Cl_{(g)}​→Cl_{(g)}^{−}​→Cl_{(aq.)}^−​\)
\(ΔH\degree=\frac{1}{2}​\times240+(−350)+(−380) \)
\(=−610\)

The correct answer is -610

Answer 2

The enthalpy change can be calculated using:

\[ \Delta H^\circ = \frac{1}{2} \times 240 + (-350) + (-380) = -610 \, \text{kJ mol}^{-1} \]

This calculation takes into account the enthalpy values of the individual components in the reaction.
The terms represent the energy associated with breaking bonds (endothermic) and forming bonds (exothermic). 
The negative value of \(\Delta H^\circ\) indicates that the reaction is exothermic, meaning it releases energy into the surroundings. 
The large magnitude of the enthalpy change suggests the reaction is highly exothermic and can drive processes such as combustion.