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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 : \(\Delta_{ dis } H _{ Cl _{2(g)}}^{\circ}=240\,kJ\,mol ^{-1}\). \(\Delta_{ eg } H _{ Cl _{(g)}}^{\circ}=-350\,kJ\,mol ^{-1}\), \(\Delta_{ hyd } H _{ Cl i _{( j )}^{\circ}}^{\circ}=-380\,kJ\,mol ^{-1}\)

  1. A \(600\)
  2. B \(620\)
  3. C \(630\)
  4. D \(610\)
Verified Solution

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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