Question

Problem 5.64 Part A 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)

Answer 1

Apply HEss Laws

inverting equation = negative change

multiplying = multiplying Hrxn

then

2C(s)+O2(g)→2CO(g)ΔH=−221.0kJ

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

then

CO(g)+2H2(g)→CH3OH(g)

invert (1)

2CO(g)→2C(s)+O2(g)ΔH= - - 221.0kJ = +221 kJ

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

divide (1) by 2 and (2) by 2

CO(g)→C(s)+1/2O2(g)ΔH= 221/2 = 110.5 kJ

C(s)+1/2O2(g)+2H2(g)→CH3OH(g)ΔH= - 201.2 kJ

add both equations

CO(g)+C(s)+1/2O2(g)+2H2(g)→C(s)+1/2O2(g) + CH3OH(g) HRxn = 110.5 - 201.2 = -90.7 kJ

cancel common terms

CO(g) +2H2(g)→ CH3OH(g) HRxn = -90.7 kJ

which is what we wanted