Answer to Question #56435 in General Chemistry for Megan

Question

Calculate ΔH for reaction 1 given reactions 2 and 3, and their respective enthalpy changes. Express your answer including units of kJ in the box below.
Reaction 1: H2(g) + CO2(g) → CO(g) + H2O(l)
Reaction 2: CO(g) + 1/2 O2(g) → CO2(g); ΔH = –283.0 kJ
Reaction 3: H2(g) + 1/2 O2(g) → H2O(l); ΔH = –285.8 kJ

Answer 1

Reaction 1: H2(g) + CO2(g) → CO(g) + H2O(l)
Reaction 2: CO(g) + 1/2 O2(g) → CO2(g); ΔH = –283.0 kJ
Reaction 3: H2(g) + 1/2 O2(g) → H2O(l); ΔH = –285.8 kJ
Reaction 1= Reaction 3 - Reaction 2
Reaction 1: H2(g) + 1/2 O2(g) + CO2(g) → H2O(l) + CO(g) + 1/2 O2(g)
H2(g) + 1/2 O2(g) + CO2(g) → H2O(l) + CO(g) + 1/2 O2(g)
ΔH(Rection 1)=ΔH(Rection 3) -ΔH(Rection 2)= –285.8+283= –2.8 kJ

Answer:
Reaction 1: H2(g) + CO2(g) → CO(g) + H2O(l); ΔH = –2,8 kJ
ΔH(Rection 1)= –2.8 kJ