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:

mathrm{cac}_{2}(s)+2 mathrm{h}_{2} mathrm{o}(g)
ightarrow mathrm{c}_{2} mathrm{h}_{2}(g)+mathrm{ca}(mathrm{oh})_{2}(s) quad delta h=-414 . mathrm{kj}

in the second step, acetylene, carbon dioxide and water react to form acrylic acid:

6 mathrm{c}_{2} mathrm{h}_{2}(g)+3 mathrm{co}_{2}(g)+4 mathrm{h}_{2} mathrm{o}(g)
ightarrow 5 mathrm{ch}_{2} mathrm{chco}_{2} mathrm{h}(g) quad delta h=132 . mathrm{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.

Explanation:

Step1: Adjust the first reaction

Multiply the first reaction by \(6\) to get \(6\) moles of \(C_2H_2\).

$$\begin{align*} 6CaC_2(s)+12H_2O(g)&\to6C_2H_2(g)+6Ca(OH)_2(s)\\ \Delta H_1&=6\times(- 414\space kJ)=-2484\space kJ \end{align*}$$

Step2: Use Hess's Law

Add the adjusted first reaction and the second reaction.

$$\begin{align*} 6CaC_2(s)+12H_2O(g)+6C_2H_2(g)+3CO_2(g)+4H_2O(g)&\to6C_2H_2(g)+6Ca(OH)_2(s)+5CH_2CHCO_2H(g)\\ 6CaC_2(s)+16H_2O(g)+3CO_2(g)&\to6Ca(OH)_2(s)+5CH_2CHCO_2H(g) \end{align*}$$

The enthalpy change \(\Delta H=\Delta H_1+\Delta H_2\), where \(\Delta H_2 = 132\space kJ\)
$$\Delta H=-2484 + 132=-2352\space kJ$$
For the formation of \(1\) mole of acrylic acid (\(CH_2CHCO_2H\)), divide the overall enthalpy change by \(5\)
$$\Delta H=\frac{-2352}{5}=-470.4\approx - 470\space kJ$$

Answer:

\(-470\space kJ\)