Discussion Forum

1)

 Given $\triangle H _{f}^{0}$  for $CO_{2} (g)$ , $CO(g)$ and $H_{2}O(g)$  are -393.5 ,-110.5 and -241.8 kJ  mol^-1, respectively . The $\triangle H_{f}^{0}$  [in  k J mol^-1] for the reaction

$CO_{2}(g)+H_{2}(g) \rightarrow  CO(g)+H_{2}O(g)$ is 

Answer:

Option D

Explanation:

$CO_{2}(g)+H_{2}(g) \rightarrow  CO(g)+H_{2}O(g)$

$\triangle H_{f}^{0}= (\triangle  H)_{products}-(\triangle H)_{reactants}$

Hence, 

$\triangle H_{f}^{0}=[\triangle H_{fCO}+\triangle H_{fH_{2}O}]-[\triangle H_{fCO_{2}}].....(i)$

 Given,   $\triangle H_{fCO_{2}(s)}=-393.5kJmol^{-1}$

$\triangle H^{0}_{fC0}=-110.5kJmol^{-1}$

$\triangle H^{0}_{fH_{2}O}=-241.8kJmol^{-1}$

Put there value in equation(i)

$\triangle H_{f}^{0}$=-110.5-241.8+393.5=41.2