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:
\ce{cac_{2}(s) + 2 h_{2}o(g) -> c_{2}h_{2}(g) + ca(oh)_{2}(s)} \quad \delta h = -414. \text{ kj}

in the second step, acetylene, carbon dioxide and water react to form acrylic acid:
\ce{6 c_{2}h_{2}(g) + 3 co_{2}(g) + 4 h_{2}o(g) -> 5 ch_{2}chco_{2}h(g)} \quad \delta h = 132. \text{ 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 \text{kj}.

Explanation:

Step1: Scale first reaction for 6 C₂H₂

Multiply first reaction by 6:
$6\text{CaC}_2(s) + 12\text{H}_2\text{O}(g)
ightarrow 6\text{C}_2\text{H}_2(g) + 6\text{Ca(OH)}_2(s)$
$\Delta H_1 = 6 \times (-414\ \text{kJ}) = -2484\ \text{kJ}$

Step2: Add second reaction to scaled first

Add scaled first and original second reaction:
$6\text{CaC}_2(s) + 12\text{H}_2\text{O}(g) + 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g)
ightarrow 5\text{CH}_2\text{CHCO}_2\text{H}(g) + 6\text{Ca(OH)}_2(s)$
$\Delta H_{\text{total}} = -2484\ \text{kJ} + 132\ \text{kJ} = -2352\ \text{kJ}$

Step3: Divide by 5 for 1 mole acrylic acid

$\Delta H_{\text{per mole}} = \frac{-2352\ \text{kJ}}{5} = -470.4\ \text{kJ} \approx -470\ \text{kJ}$

Answer:

-470