JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
The enthalpy change for the conversion of \(\frac{1}{2} Cl _2\) (g) to \(Cl ^{-}\)(aq) is \((-).......\) \(kJ\,mol ^{-1}\) (Nearest integer) Given : \(\Delta_{ dis } H _{ Cl _{2(g)}}^{\circ}=240\,kJ\,mol ^{-1}\). \(\Delta_{ eg } H _{ Cl _{(g)}}^{\circ}=-350\,kJ\,mol ^{-1}\), \(\Delta_{ hyd } H _{ Cl i _{( j )}^{\circ}}^{\circ}=-380\,kJ\,mol ^{-1}\)
- A \(600\)
- B \(620\)
- C \(630\)
- D \(610\)
Answer & Solution
Correct Answer
(D) \(610\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{2} Cl _{2( g )} \rightarrow Cl _{( g )} \rightarrow Cl _{( g )}^{-} \rightarrow Cl _{( aq .)}^{-}\) \(\Delta H ^{\circ}=\frac{1}{2} \times 240+(-350)+(-380)\) \(=-610 \text { ans. }\)
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