Question

Manipulate equations to solve enthalpy: 2c2h2(g) 5o2(g)⟶4co2(g) 2h2o(g) 2c2h2(g) 5o2(g)⟶4co2(g) 2h2o(g) compound δh∘f (kj/mol)δhf∘ (kj/mol) c2h2(g)c2h2(g) 227.4227.4 o2(g)o2(g) 00 co2(g)co2(g) −393.5−393.5 h2o(g)h2o(g) −241.8−241.8 what is the δh∘δh∘ of the reaction?

Answer 1

The enthalpy change (Delta H degrees) of the given reaction is calculated using the standard enthalpies of formation and comes out to be -2512.4 kJ.

Step-by-step explanation:

To determine the ΔH° of the reaction 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g), we will use the provided standard enthalpies of formation (ΔH°f) for each compound. The standard enthalpy change of a reaction can be calculated using the following formula:

ΔH°rxn = ∑(ΔH°f,products) − ∑(ΔH°f,reactants)

First, we sum up the enthalpies of formation for the products (4 moles of CO2 and 2 moles of H2O) and then subtract the sum of the enthalpies of formation for the reactants (2 moles of C2H2 and 5 moles of O2). Given that the ΔH°f for O2(g) is zero, it will not contribute to the calculation.

Using the provided data:

ΔH°rxn = [4(-393.5 kJ/mol) + 2(-241.8 kJ/mol)] − [2(227.4 kJ/mol) + 5(0 kJ/mol)]

ΔH°rxn = [-1574 kJ + (-483.6 kJ)] − [454.8 kJ + 0 kJ]

ΔH°rxn = -2057.6 kJ − 454.8 kJ = -2512.4 kJ

The enthalpy change for this reaction is -2512.4 kJ.