Questions: 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) + 2 H2O(g) → C2H2(g) + Ca(OH)2(s) ΔH = -414 kJ In the second step, acetylene, carbon dioxide and water react to form acrylic acid: 6 C2H2(g) + 3 CO2(g) + 4 H2O(g) → 5 CH2CHCO2H(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.

Solution Steps

The first reaction is: \[ \mathrm{CaC}_{2}(\mathrm{~s}) + 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) + \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{~s}) \quad \Delta H = -414 \, \mathrm{kJ} \]

This reaction produces 1 mole of acetylene (\(\mathrm{C}_{2}\mathrm{H}_{2}\)) with an enthalpy change of \(-414 \, \mathrm{kJ}\).

The second reaction is: \[ 6 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) + 3 \mathrm{CO}_{2}(\mathrm{~g}) + 4 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \rightarrow 5 \mathrm{CH}_{2} \mathrm{CHCO}_{2} \mathrm{H}(\mathrm{~g}) \quad \Delta H = 132 \, \mathrm{kJ} \]

This reaction produces 5 moles of acrylic acid (\(\mathrm{CH}_{2}\mathrm{CHCO}_{2}\mathrm{H}\)) with an enthalpy change of \(132 \, \mathrm{kJ}\).

The enthalpy change per mole of acrylic acid is: \[ \Delta H_{\text{per mole of acrylic acid}} = \frac{132 \, \mathrm{kJ}}{5} = 26.4 \, \mathrm{kJ} \]

To form one mole of acrylic acid, we need:

• 1 mole of acrylic acid from the second reaction, which requires \(\frac{6}{5}\) moles of acetylene.
• The enthalpy change for producing \(\frac{6}{5}\) moles of acetylene from the first reaction.

The enthalpy change for producing \(\frac{6}{5}\) moles of acetylene is: \[ \Delta H_{\text{acetylene}} = \left(\frac{6}{5}\right) \times (-414 \, \mathrm{kJ}) = -496.8 \, \mathrm{kJ} \]

The net enthalpy change for the formation of one mole of acrylic acid is the sum of the enthalpy changes from both steps: \[ \Delta H_{\text{net}} = -496.8 \, \mathrm{kJ} + 26.4 \, \mathrm{kJ} = -470.4 \, \mathrm{kJ} \]

Final Answer