The enthalpy change for the reaction 5 C(s) + 6 H₂(g) + 1/2 O₂ (g) → C₅H₁₁OH(l) is ΔH = -566.0 kJ/mol.
We can solve this problem using Hess's Law and the data provided.
By manipulating the given equations, we can obtain the desired reaction equation:
2 C(s) + O₂(g) → 2 CO(g), ΔH = +221.0 kJ
C(s) + O₂(g) → CO₂(g), ΔH = -393.5 kJ
Multiplying the second equation by 2, we get:
2 C(s) + 2 O₂(g) → 2 CO₂(g), ΔH = -787.0 kJ
Now we can add these two equations:
2 C(s) + O₂(g) → 2 CO(g), ΔH = +221.0 kJ
2 C(s) + 2 O₂(g) → 2 CO₂(g), ΔH = -787.0 kJ
-----------------------------
O₂(g) → CO(g), ΔH = -566.0 kJ
Therefore, the enthalpy change for the reaction 5 C(s) + 6 H₂(g) + 1/2 O₂ (g) → C₅H₁₁OH(l) is ΔH = -566.0 kJ/mol.
The probable question may be:
Calculate Δ_T H^* for the reaction
5 C(s) + 6 H_2(g) + 1/2 O_2 (g) → C5 H_{11} OH(l)
given the information below.
C(s) + O_2(g) → CO_2 (g) Δ_{r1}H^*=-393.5 kJ/mol-rxn
2 H_2(g) + O_2(g) → 2 H_2O(l) Δ_{r2} H^*=571.6 kJ/mol-rxn
C5H11OH() + 15/2 O2 (g) → 5CO2 (g) + 6H2O(l) Δ_{r3} H^*=-3330.7 kJ/mol-rxn
Δ_{r} H^*=___________kJ/mol-rxn