Question

Calculate ΔH° for the reaction C4H4 (g) + 2H2 (g) ---> C4H8 (g) using the following date:
ΔH° combustion [C4H4 (g)] = -2341 kJ/mole
ΔH° combustion [H2 (g)] = -286 kJ/mole
ΔH° combustion [C4H8 (g)] =-2755 kJ/mole

a. 128 kJ
b. 158 kJ
c. -158 kJ
d. -128kJ
e. none given

Answer 1

The enthalpy change, ΔH° for the reaction is; 158 kJ/mol

The enthalpy change for the reaction can be evaluated as follows;

ΔH° = H°(products) - H°(reactants)

• where; H°(products) = (1 × -2755 kJ/mol) = -2755 kJ/mol

• H°(reactants) = (1 × -2341 kJ/mol) + (2× -286kJ/mol) = -2913 kJ/mol.

• ΔH° = -2755 - (-2913)

• ΔH° = 158 kJ/mol

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