Calculate ΔHrxn for the following reaction: C(s) + H2O(g) --> CO(g) + H2(g) Use the following reactions and given ΔH values: C (s) + O2 (g) → CO2 (g), ΔH = -393.5 kJ 2 CO (g) + O2 (g) → 2 CO2 (g), ΔH= -566.0 kJ 2 H2 (g) + O2 (g) → 2 H2O (g), ΔH= -483.6 kJ Express your answer numerically, to four significant figures and in terms of kJ.

ΔH = 130.5 kJ

Explanation:

Hello,

In this case, by using the Hess law, we compute the enthalpy of the required reaction:

C(s) + H2O(g) --> CO(g) + H2(g)

Thus, the first step is to keep the following reaction unchanged:

C (s) + O2 (g) → CO2 (g), ΔH = -393.5 kJ

Next, we invert and halve this reaction:

2 CO (g) + O2 (g) → 2 CO2 (g), ΔH= -566.0 kJ

So the enthalpy of reaction is inverted and halved:

CO2 (g) → CO (g) + 1/2 O2 (g) ΔH= 283 kJ

Then, we also invert and halve this reaction:

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

So the enthalpy of reaction is inverted and halved as well:

H2O → H2 (g) + 1/2 O2 (g) ΔH= 241.8 kJ

Finally, we add the three reactions to obtain the required reaction:

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

= C (s) + O2 (g) + CO2 (g) + H2O → H2 (g) + O2 (g) + CO (g) + CO2 (g)

= C (s) + H2O → H2 (g) CO (g)

So enthalpy is computed by:

ΔH = -393.5 kJ + 283 kJ + 241.8 kJ

ΔH = 130.5 kJ

Best regards.

b.final

28528289 29823282 8298232 39858