Calculate AH for the reaction CH4 (g) + NH3 (g) → HCN (g) + 3 H2 (g), from the reactions.
N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔΗ = -91.8 kJ
C (s, graphite) + 2 H2 (g) → CH4 (g) ΔΗ = -74.9 kJ/mole
H2 (g) + 2 C (s, graphite) + N2 (g) → 2 HCN (g) ΔΗ = +270.3 kJ
We can use the given reactions to obtain the overall reaction and calculate the enthalpy change (AH).
1. Multiply the given reaction 1 by -1 in order to reverse it:
-1(N2 (g) + 3 H2 (g) → 2 NH3 (g))
gives:
-N2 (g) - 3 H2 (g) → -2 NH3 (g)
2. Multiply the given reaction 2 by 3 in order to obtain 3 H2 (g):
3(C (s, graphite) + 2 H2 (g) → CH4 (g))
gives:
3 C (s, graphite) + 6 H2 (g) → 3 CH4 (g)
3. Add the above two reactions together:
-N2 (g) - 3 H2 (g) + 3 C (s, graphite) + 6 H2 (g) → -2 NH3 (g) + 3 CH4 (g)
Simplifying the equation and cancelling the like terms, we get:
2 C (s, graphite) + 3 H2 (g) → -2 NH3 (g) + 3 CH4 (g)
Now we also have the given reaction:
H2 (g) + 2 C (s, graphite) + N2 (g) → 2 HCN (g)
To obtain the overall reaction, we can add this reaction to the equation derived in Step 3:
2 C (s, graphite) + 3 H2 (g) + H2 (g) + 2 C (s, graphite) + N2 (g) → -2 NH3 (g) + 3 CH4 (g) + 2 HCN (g)
Simplifying the equation and cancelling the like terms, we get:
4 C (s, graphite) + 4 H2 (g) + N2 (g) → -2 NH3 (g) + 3 CH4 (g) + 2 HCN (g)
Now we can calculate the enthalpy change (AH) using the given values:
ΔΗ = ΣΔΗ(products) - ΣΔΗ(reactants)
ΔΗ = (-2 NH3 (g) + 3 CH4 (g) + 2 HCN (g)) - (4 C (s, graphite) + 4 H2 (g) + N2 (g))
ΔΗ = (-2 × (-91.8 kJ) + 3 × (-74.9 kJ/mole) + 2 × (+270.3 kJ)) - (4 × 0 + 4 × 0 + 0)
ΔΗ = 183.6 kJ + (-224.7 kJ) + 540.6 kJ
ΔΗ = 499.5 kJ
Therefore, the enthalpy change (AH) for the reaction CH4 (g) + NH3 (g) → HCN (g) + 3 H2 (g) is 499.5 kJ.