Answer:
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) + <u>1/2 O2</u> (g) + CO (g) + <u>1/2 O2 (g)</u> + CO2 (g)
= C (s) + <u>O2 (g)</u> + <u>CO2 (g) </u>+ H2O → H2 (g) + <u>O2 (g)</u> + CO (g) + <u>CO2 (g)</u>
= 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.
