Sure, let's work through the problem step-by-step:
### Step 1: Understand the given reactions and their enthalpy changes
1. First Reaction: Formation of acetylene from calcium carbide and water:
[tex]\[
CaC_2(s) + 2 H_2O(g) \rightarrow C_2H_2(g) + Ca(OH)_2(s) \quad \Delta H = -414 \, \text{kJ}
\][/tex]
2. Second Reaction: Formation of acrylic acid from acetylene, carbon dioxide, and water:
[tex]\[
6 C_2H_2(g) + 3 CO_2(g) + 4 H_2O(g) \rightarrow 5 CH_2CHCO_2H(\ell) \quad \Delta H = 132 \, \text{kJ}
\][/tex]
### Step 2: Determine how many moles of acrylic acid are formed in the second reaction
In the second reaction, 5 moles of acrylic acid ([tex]\(CH_2CHCO_2H\)[/tex]) are formed.
### Step 3: Calculate the total enthalpy change for the formation of acrylic acid
The objective is to determine the net enthalpy change for the formation of one mole of acrylic acid. First, we sum the enthalpy changes for the given reactions:
[tex]\[
\Delta H_{\text{total}} = \Delta H_1 + \Delta H_2
\][/tex]
Given:
[tex]\[
\Delta H_1 = -414 \, \text{kJ}, \quad \Delta H_2 = 132 \, \text{kJ}
\][/tex]
Thus:
[tex]\[
\Delta H_{\text{total}} = -414 \, \text{kJ} + 132 \, \text{kJ} = -282 \, \text{kJ}
\][/tex]
### Step 4: Divide the total enthalpy change by the number of moles of acrylic acid formed to find the enthalpy change per mole
Since 5 moles of acrylic acid are formed, the enthalpy change per mole of acrylic acid is calculated as:
[tex]\[
\Delta H_{\text{per mole}} = \frac{\Delta H_{\text{total}}}{\text{moles of acrylic acid}} = \frac{-282 \, \text{kJ}}{5}
\][/tex]
### Step 5: Perform the division
[tex]\[
\Delta H_{\text{per mole}} = \frac{-282}{5} = -56.4 \, \text{kJ/mol}
\][/tex]
### Step 6: Round to the nearest kJ
[tex]\[
-56.4 \, \text{kJ/mol} \approx -56 \, \text{kJ/mol}
\][/tex]
### Final Answer
The net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water, and carbon dioxide is -56 kJ/mol, rounded to the nearest kJ.
