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Calculate delta H(r)^(0) for the reactio...

Calculate `delta H_(r)^(0)` for the reaction
`CO_(2)(g) + H_(2)(g) rarr CO(g) + H_(2)O(g)`
given that `Delta H_(r)^(0)` for `CO_(2)` (g), CO (g) and `H_(2)O` (g) are `- 393.5, -111.31` and `- 242 KJ mol^(-1)` respectively.

Text Solution

Verified by Experts

Given: `Delta H_(f)^(0) CO_(2) = -393.5kJ mol^(-1)`
`DeltaH_(f)^(0) CO = -111.31kJ mol^(-1)`
`DeltaH_(f)^(0) (H_(2)O) = -242kJ mol^(-1)`
`CO_(2)(g) + H_(2)(g) rarr CO(g) + H_(2)O(g)`
`DeltaH_(r)^(0) =?`
`DeltaH_(r)^(0) = Sigma(DeltaH_(f)^(0))_("products")- Sigma (DeltaH_(f)^(0))_("reactants")`
`DeltaH_(r)^(0) = [DeltaH_(f)^(0) (CO) + DeltaH_(f)^(0) (H_(2)O)]- [DeltaH_(f)^(0) (CO_(2)) + DeltaH_(f)^(0) (H_(2))]`
`DeltaH_(r)^(0) = [-111.31 + (-242)] -[-393.5+ (0)]`
`Delta H_(r)^(0) = [-353.31] + 393.5`
`DeltaH_(r)^(0) = 40.19`
`DeltaH_(r)^(0) = +40.19 kJ mol^(-1)`
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