Question

using enthalpies of formation, determine the balanced thermal chemical equation for the following reaction:
2 c₂h₆ (g) + 2 ccl₄ (g) → 3 c₂h₄cl₂ (l) + cl₂ (g)
how much heat is given off or absorbed when 70.42 g of c₂h₄cl₂ is formed. the reaction occurs at 25.0°c at a constant pressure of 1.00 atm. note that the enthalpies of formation can be found in the tables of chemical data posted on canvas. report the answer in kilojoules to 4 sig figs, but do not include units in your answer.

Explanation:

Step1: Find Molar Mass of \( C_2H_4Cl_2 \)

Molar mass of \( C_2H_4Cl_2 \): \( 2\times12.01 + 4\times1.008 + 2\times35.45 = 98.956 \, \text{g/mol} \)

Step2: Calculate Moles of \( C_2H_4Cl_2 \)

Moles \( = \frac{70.42 \, \text{g}}{98.956 \, \text{g/mol}} \approx 0.7116 \, \text{mol} \)

Step3: Recall Enthalpy of Reaction Formula

\( \Delta H_{rxn} = \sum n\Delta H_f(\text{products}) - \sum n\Delta H_f(\text{reactants}) \)
From data tables:
\( \Delta H_f(C_2H_6(g)) = -84.68 \, \text{kJ/mol} \), \( \Delta H_f(CCl_4(g)) = -135.44 \, \text{kJ/mol} \)
\( \Delta H_f(C_2H_4Cl_2(l)) = -165.2 \, \text{kJ/mol} \), \( \Delta H_f(Cl_2(g)) = 0 \, \text{kJ/mol} \)

Step4: Calculate \( \Delta H_{rxn} \) for 3 mol \( C_2H_4Cl_2 \)

\( \sum n\Delta H_f(\text{products}) = 3\times(-165.2) + 1\times0 = -495.6 \, \text{kJ} \)
\( \sum n\Delta H_f(\text{reactants}) = 2\times(-84.68) + 2\times(-135.44) = -2\times84.68 - 2\times135.44 = -169.36 - 270.88 = -440.24 \, \text{kJ} \)
\( \Delta H_{rxn} = -495.6 - (-440.24) = -55.36 \, \text{kJ} \) (for 3 mol \( C_2H_4Cl_2 \))

Step5: Find \( \Delta H \) for 0.7116 mol \( C_2H_4Cl_2 \)

Per 3 mol \( C_2H_4Cl_2 \), \( \Delta H = -55.36 \, \text{kJ} \)
For 0.7116 mol: \( \Delta H = \frac{-55.36 \, \text{kJ}}{3 \, \text{mol}} \times 0.7116 \, \text{mol} \approx -13.07 \)

Answer:

-13.07