working out enthalpy change for C(s) + 2H2O(g) ⇌ CO2(g) + 2H2(g)

Question

C(s) + 2H2O(g) ⇌ CO2(g) + 2H2(g) work out the enthalpy change for this equation with the following information.

C(s) + O2(g) → CO2(g) =-393.5kjmol-1

H2(g) + ½ O2(g) → H2O(g)= -241.8 kjmol=1

Answer 1

Write the given chemical equations like this:

C(s) + O2(g) → CO2(g) ΔH = -393.5 kjmol-1

H2O(g) → H2(g) + ½ O2(g) ΔH = + 241.8 kjmol-1 ( Note the sign change for ΔH since the given equation is written in reverse.

Multiply the second equation by 2 (including the ΔH). This gives:

C(s) + O2(g) → CO2(g) ΔH = -393.5 kjmol-1

2H2O(g) → 2H2(g) + O2(g) ΔH = + 483.6 kjmol-1

Add the last two equations, cancelling out like compounds on each side.

C(s) + O2(g) + 2H2O(g) → CO2(g) + 2H2(g) + O2(g) ΔH = -393.5 kjmol-1 + 483.6 kjmol-1 - = + 90.1 kjmol-1 .

The last equation simplifies to the original equation given in the problem. This procedure is known as Hess's law.

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