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(g) → 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) =ΔH132.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

Answer:

The net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide is 523.2 kJ.

Explanation:

Step 1:

$CaC_2(s) + 2H_2O(g)\rightarrow C_2H_2(g) + Ca(OH)_2(s),\Delta H_1=414.0 kJ$...[1]

Step 2 :

$6C_2H_2(g) + 3CO_2(g) + 4H_2O(g)\rightarrow 5CH_2CHCO_2H(g) \Delta H_2=132.0kJ$..[2]

Adding 6 × [1] and [2]:

$6CaC_2(s) + 12H_2O(g)\rightarrow 6C_2H_2(g) + 6Ca(OH)_2(s)$

$6C_2H_2(g)+3CO_2(g)+16H_2O(g)\rightarrow 5CH_2CHCO_2H(g)$

we get :

$6CaC_2(s) + 8H_2O(g)+3CO_2(g)\rightarrow 5CH_2CHCO_2H(g)+ 6Ca(OH)_2(s),\Delta H'=?$

$\Delta H'=6\times \Delta H_1+\Delta H_2$

$\Delta H'=6\times 414.0 kJ+132.0kJ$

$\Delta H'=2,626 kJ$

Energy released on formation of 5 moles of acrylic acid = 2,626 kJ

Energy released on formation of 1 mole of acrylic acid:

$\frac{ 2,626 kJ}{5 } = 523.2 kJ$