The enthalpy change for the conversion of $\dfrac{1}{2}{Cl}_{2}$ (g) to ${Cl}^{-} ( aq )$ is $( - )$
${kJmol}^{- 1}$ (Nearest integer)
Given : ${\Delta}_{\text{dis}}{H}_{{Cl}_{2 ( g )}}^{0} = 240{kJmol}^{- 1}$ .
${\Delta}_{eg}{H}_{Cl ( g )}^{o} = - 350{kJmol}^{- 1}$ ,
${\Delta}_{hyd}{H}_{{Cl}_{( g )}^{-}}^{o} = - 380{kJmol}^{- 1}$

Question

The enthalpy change for the conversion of

\dfrac{1}{2}{Cl}_{2}

(g) to

{Cl}^{-} ( aq )

is

( - )

{kJmol}^{- 1}

(Nearest integer)
Given :

{\Delta}_{\text{dis}}{H}_{{Cl}_{2 ( g )}}^{0} = 240{kJmol}^{- 1}

.

{\Delta}_{eg}{H}_{Cl ( g )}^{o} = - 350{kJmol}^{- 1}

,

{\Delta}_{hyd}{H}_{{Cl}_{( g )}^{-}}^{o} = - 380{kJmol}^{- 1}

Quantrex Academy · Accepted Answer

Correct answer: 0

(1)/(2) Cl _ 2 ( g ) → Cl _ ( g ) → Cl _ ( g ) ^ - → Cl _ ( aq . ) ^ - Apply Hess's law, ∆ H = (1)/(2) ∆ _ dis H _ Cl _ 2 ( g ) ^ 0 + ∆ _ eg H _ Cl ( g ) ^ o + ∆ _ hyd H _ Cl _ ( g ) ^ - ^ o ⇒ ∆ H = (1)/(2) × 240 + ( - 350 ) + ( - 380 ) ⇒ ∆ H = - 610

The enthalpy change for the conversion of (1)/(2) Cl _ 2 (g) to Cl...

Correct answer

Step-by-step solution

More from Thermodynamics (C)