Enthalpies of formation of CCl 4 ( g ) , H 2 O ( g ) , CO 2 ( g ) and HCl are - 105 , - 242 , - 394 and - 92 kJ mol - 1 respectively. The magnitude of enthalpy of the reaction given below is kJmol - 1

Question

Enthalpies of formation of CCl 4 ( g ) , H 2 O ( g ) , CO 2 ( g ) and HCl are - 105 , - 242 , - 394 and - 92 kJ mol - 1 respectively. The magnitude of enthalpy of the reaction given below is kJmol - 1 . (nearest integer) CCl 4 ( g ) + 2 H 2 O ( g ) → CO 2 ( g ) + 4 HCl ( g )

TestHub · Accepted Answer

Using Hess's law of constant heat summation:
Enthalpy of reaction = Enthalpy of formation of products - Enthalpy of formation of reactants
$∴ {ΔH}_{f} ° = {ΔH}_{f} ° ({CO}_{2}, g) + 4 {ΔH}_{f} ° (HCl) - {ΔH}_{f} ° ({CCl}_{4}) - 2 {ΔH}_{f} ° (H_{2} O, r)$

$= [- 394] + 4 [- 92] + 105 - 2 \times [242]$

$= - 394 - 368 + 105 + 484$

$= - 173 kJ / mole$

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