If the enthalpy of formation and enthalpy of solution of HCl ( g ) are − 90 KJ mol − 1 and − 75 KJ mol − 1 then enthalpy of formation of Cl ( aq ) − is [ ( Δ f H ∘ ) H ( aq ) + = 0 ]

Question

If the enthalpy of formation and enthalpy of solution of HCl ( g ) are − 90 KJ mol − 1 and − 75 KJ mol − 1 then enthalpy of formation of Cl ( aq ) − ​ is [ ( Δ f ​ H ∘ ) H ( aq ) + ​ = 0 ]

TestHub · Accepted Answer

Question Explanation: Calculating enthalpy of formation for $Cl^{-} (aq)$ .

Concept: Enthalpy of Formation. Solution:

The enthalpy of solution of HCl(g) equals the difference between the enthalpies of formation of its aqueous products (H⁺, Cl⁻) and the gaseous reactant.

Calculation:

$\frac{1}{2} H_{2} + \frac{1}{2} C l_{2} \to H Cl (1)$

$Δ_{f} H = - 90 k J m o l^{- 1}$

$H Cl \to H^{+} + C l^{-} (2)$

$Δ_{so l} H = - 75 k J m o l^{- 1}$

Adding equations (1) and (2):

$\frac{1}{2} H_{2} + \frac{1}{2} C l_{2} \to H^{+} + C l^{-}$

$Δ_{f} H = (- 90) + (- 75)$

$= - 165 k J m o l^{- 1}$ because $(Δ_{f} H^{0})_{H^{+} (a q)} = 0$ .

Final Answer: -165

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