Using the table below that shows some values for standard enthalpy changes of combustion, calculate the standard enthalpy change of the reaction, C(s) + 2H2(g) rightwards arrowCH4(g)
Substance
ΔH°f (kJ/mol)
C(s)
−394
CH4(g)
−890
H2(g)
−286
-2225 kJ/mol
-470 kJ/mol
+2225 kJ/mol
+470kJ/mol
+470 kJ/mol
To calculate the standard enthalpy change of the reaction, we need to use the Hess's Law, which states that the enthalpy change of a reaction is the same regardless of the route taken.
The balanced equation for the reaction is:
C(s) + 2H2(g) → CH4(g)
We can break down the reaction into two steps:
Step 1: C(s) → CO(g) + O2(g)
Step 2: CO(g) + 2H2(g) → CH4(g)
Now, let's calculate the enthalpy change for each step:
Step 1: The enthalpy change for this step is the standard enthalpy of formation (ΔH°f) of CO(g) minus the standard enthalpy of formation of C(s) and O2(g), since they are the reactants.
ΔH1 = ΔH°f(CO(g)) - ΔH°f(C(s)) - 1/2 ΔH°f(O2(g))
ΔH1 = 0 - (-394) - 1/2(0) = 394 kJ/mol
Step 2: The enthalpy change for this step is the standard enthalpy of formation of CH4(g) minus the standard enthalpy of formation of CO(g) and 2H2(g), since they are the reactants.
ΔH2 = ΔH°f(CH4(g)) - ΔH°f(CO(g)) - 2ΔH°f(H2(g))
ΔH2 = (-890) - 0 - 2(-286) = -890 + 572 = -318 kJ/mol
Finally, we can calculate the overall enthalpy change for the reaction by summing the enthalpy changes for each step:
Overall ΔH = ΔH1 + ΔH2 = 394 kJ/mol + (-318 kJ/mol) = +76 kJ/mol
Therefore, the standard enthalpy change of the reaction C(s) + 2H2(g) → CH4(g) is +76 kJ/mol.