Question

From the following data, C(graphite) + O2(g) ⟶CO2(g) ΔH°rxn = −393.5 kJ/mol H2(g) + 12 O2(g) ⟶H2O(l) ΔH°rxn = −285.8 kJ/mol 2C2H6(g) + 7O2(g) ⟶4CO2(g) + 6H2O(l) ΔH°rxn = −3119.6 kJ/mol calculate the enthalpy change for the reaction 2C(graphite) + 3H2(g) ⟶C2H6(g

Answer 1

The enthalpy change for the reaction 2C(graphite) + 3H₂(g) ⟶C₂H₆(g) is -6018.2 kJ/mol.

Step-by-step explanation:

The enthalpy change for the reaction 2C(graphite) + 3H₂(g) ⟶C₂H₆(g) can be calculated using Hess's law and the given enthalpy changes for the provided reactions.

We need to manipulate the given reactions and combine them in a way that cancels out the desired reactants and products. By reversing the first reaction and multiplying the second reaction by 2, we can obtain the desired equation.

The enthalpy change for the reversed first reaction is +221.0 kJ, and the enthalpy change for the second reaction multiplied by 2 is -6239.2 kJ. Adding these two equations together gives a balanced equation with the desired reactants and products, and the sum of the enthalpy changes is -6018.2 kJ/mol.