Question

There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In the first step, calcium carbide and water react to form acetylene and calcium hydroxide:
CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(s)ΔH = −414kJ
In the second step, acetylene, carbon dioxide and water react to form acrylic acid:
6C2H2(g) + 3CO2(g) + 4H2O(g) → 5CH2CHCO2H(g)ΔH = 132kJ
Calculate the net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water, and carbon dioxide from these reactions. Round your answer to the nearest kJ.

Answer 1

Answer:

ΔH = -470.4kJ

Explanation:

It is possible to sum 2 or more reactions to obtain the ΔH of the reaction you want to study (Hess's law). Using the reactions:

1. CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(s)ΔH = −414kJ

2. 6C2H2(g) + 3CO2(g) + 4H2O(g) → 5CH2CHCO2H(g)ΔH = 132kJ

6 times the reaction 1.

6CaC2(s) + 12H2O(l) → 6C2H2(g) + 6Ca(OH)2(s)ΔH = −414kJ*6 = -2484kJ

This reaction + 2:

6CaC2(s) + 3CO2(g) + 16H2O(l) →  + 6Ca(OH)2(s) + 5CH2CHCO2H(g) ΔH = -2484kJ + 132kJ = -2352kJ

As we want to calculate the net change enthalpy in the formation of just 1 mole of acrylic acid we need to divide this last reaction in 5:

6/5CaC2(s) + 3/5CO2(g) + 16/5H2O(l) →  + 6/5Ca(OH)2(s) + CH2CHCO2H(g) ΔH = -2352kJ / 5

ΔH = -470.4kJ