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Calculate the enthalpy of formation of C...

Calculate the enthalpy of formation of `CH_3 OH (l) + 3/2 O_2 (g) to CO_2(g) + 2H_2O (l) " " Delta H^@ = - 726 kJ "mol"^(-1)`
`C(s) + O_2 (g) to CO_2 (g) " " Delta H^@ = -393 kJ "mol"^(-1)`
`H_2 (g) + 1/2 O_2 (g) to H_2O (l) " " Delta H^@ = -286 kJ "mol"^(-1)`

Text Solution

Verified by Experts

The required equation is
`C(s) + 2H_2 (g) + 1/2 O_2 (g) to CH_3 OH (l) Delta H = ? `
The given equations can be rearranged as
(i) `CO_2(g) + 2H_2 O(l) to CH_3OH + 3/2 O_2 (g) + 726 KJ "mol"^(-1)`
(ii) `C(s) + O_2(g) to CO_2 (g) -393 KJ mol^(-1)`
(iii) `2H_2 (g) + O_2 (g) to 2H_2 O (l) 2 xx -286 KJ mol^(-1)`
On adding `C(s) + 2H_2(g) + 1/2 O_2 (g) to CH_3 OH (l) - 239.0 KJ`
`DeltaH^@ (CH_3 OH) = -239.0 kj mol^(-1)`
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