1. Tardigrade
  2. Question
  3. Chemistry
  4. The ΔH°f for CO2(g) , CO (g) and H2O (g) are −393.5 , −110.5 and −241.8 kJ/mol respectively. The standard enthalpy change (in kJ ) for the reaction CO2(g) + H2(g) longrightarrow CO (g) + H2O (g) is :

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Q. The $ ΔH^°_f $ for $ CO_2(g) $ , $ CO\,(g) $ and $ H_2O\, (g) $ are $ −393.5 $ , $ −110.5 $ and $ −241.8\, kJ/mol $ respectively. The standard enthalpy change (in $ kJ $ ) for the reaction
$ CO_2(g) + H_2(g) \longrightarrow CO \,(g) + H_2O\, (g) $ is :

UPSEEUPSEE 2005

Solution:

$ \Delta \,H=\Sigma\left(\Delta H^{\circ} \text { products }\right)-\Sigma\left(\Delta H^{\circ} \text { reactants }\right)$

Given, $ \Delta \,H_{f}^{\circ} CO _{2}(g)=-393.5 \,kJ / mol $

$ \Delta \,H_{f}^{\circ} CO (g)=-110.5 kJ / mol $

$ \Delta \,H^{\circ} f H _{2} O (g)=-241.8 kJ / mol $

$ CO _{2}(g)+ H _{2}(g) \longrightarrow CO (g)+ H _{2} O (g) $

$\Delta H^{\circ}=\left[\left(\Delta H^{\circ}{ }_{f} CO \right)+\left(\Delta H ^{\circ} f H _{2} O \right)\right] $

$-\left[\left(\Delta H ^{\circ}{ }_{f} CO _{2}\right)+\left(\Delta H ^{\circ}{ }_{f} H _{2}\right)\right] $

$=[(-110.5)+(-241.8)]$

$-[(-393.5)+(0)]$

$=-110.5-241.8+393.5$

$=-352.3+393.5 $

$=+41.2\, kJ $