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) + CaOH2 (s) ΔH=−414.kJ In the second step, acetylene, carbon dioxide and water react to form acrylic acid: 6C2H2 (g) + 3CO2 (g) + 4H2O (g) → 5CH2CHCO2H (g) ΔH=132.kJ 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

Solution :-

Balanced reactions equations are as follows.

CaC2 (s) + 2H2O (l) → C2H2 (g) + CaOH2 (s) ΔH=−414.kJ

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

To get the 6 C2H2 we need to multiply first equation by 6

Then we get

6CaC2 (s) + 12H2O (l) → 6C2H2 (g) + 6CaOH2 (s)        ΔH=−414.kJ * 6 = -2484 kJ

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

So the total energy produced = -2484 kJ + 132.0 kJ = -2352 kJ

This amount of energy is given out for the 5 moles of acid

So lets convert it for 1 mol

-2352 kJ * 1 mol / 5 mol = -470.4 kJ/mol

So the delta H of reaction for the formation of the acrylic acid is -470.4 kJ/mol