Question

Calculate ΔrH for the following reaction: C(s)+H2O(g)→CO(g)+H2(g)

Use the following reactions and given ΔrH's.

C(s)+O2(g)→CO2(g), ΔrH= -393.5 kJmol−1

2CO(g)+O2(g)→2CO2(g), ΔrH= -566.0 kJmol−1

2H2(g)+O2(g)→2H2O(g), ΔrH= -483.6 kJmol−1

Express your answer using one decimal place.

Answer 1

Calculate ΔrH for the following reaction: C(s)+H2O(g)→CO(g)+H2(g)

Use the following reactions and given ΔrH's.

C(s)+O2(g)→CO2(g), ΔrH= -393.5 kJmol−1

2CO(g)+O2(g)→2CO2(g), ΔrH= -566.0 kJmol−1

2H2(g)+O2(g)→2H2O(g), ΔrH= -483.6 kJmol−1

eq.1 C(s)+O2(g)→CO2(g), ΔrH= -393.5 kJmol−1

eq.2 2CO(g)+O2(g)→2CO2(g), ΔrH= -566.0 kJmol−1

eq.3 2H2(g)+O2(g)→2H2O(g), ΔrH= -483.6 kJmol−1

To get given eq. we have to do the following [eq.1 - 1/2(eq.2) -1/2(eq.3)] =

C(s)+O2(g) - CO(g) - 1/2O2(g) - H2(g) -1/2O2(g) → CO2(g) - CO2(g) –H2O(g)

Or C(s)+ H2O(g) → CO(g) + H2(g)

ΔrH = (-393.5 kJmol−1) –1/2(-566.0 kJmol−1) – ½(-483.6 kJmol−1) = 131.3 kJmol−1