From the enthalpies of reaction

Question

From the enthalpies of reaction
2C(s)+O2(g)→2CO(g)ΔH=−221.0kJ
2C(s)+O2(g)+4H2(g)→2CH3OH(g)ΔH=−402.4kJ
calculate ΔH for the reaction
CO(g)+2H2(g)→CH3OH(g)
Express your answer with the appropriate units.

Answer 1

To answer this question, we will be using Hess's Law (look it up if you are not familiar with it)

CO(g) + 2H2(g)→CH3OH(g) ... TARGET EQUATION

Given:

(1). 2C(s) + O2(g)→2CO(g) ΔH=−221.0kJ

(2). 2C(s) + O2(g) + 4H2(g)→2CH3OH(g) ΔH=−402.4kJ

Procedure:

reverse eq.1 and divide by 2: CO(g) --> C(s) + 1/2 O2(g) ... ∆H = +110.5 kJ

divide eq 2 by 2: C(s) + 1/2 O2(g) + 2H2(g) --> CH3OH(g) ... ∆H = -201.2 kJ

add the two equations: CO(g) + C(s) + 1/2 O2(g) + 2H2(g) --> C(s) + 1/2 O2(g) + CH3OH(g)

cancel like terms: COg(g) + 2H2(g) -- CH3OH(g) ... This is the target equation

add up the ∆G values: 110.5 kJ -201.2 kJ = -90.7 kJ (answer)