For the reaction \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}(1)\) determine \(\Delta H^{\circ},\) given that $$\begin{array}{r} 4 \mathrm{HCl}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{Cl}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(1) \\ \Delta H^{\circ}=-202.4 \mathrm{kJ} \end{array}$$ $$\begin{aligned} 2 \mathrm{HCl}(\mathrm{g})+\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \\ \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}(1)+\mathrm{H}_{2} \mathrm{O}(1) & \Delta H^{\circ}=-318.7 \mathrm{kJ} \end{aligned}$$

Step by step solution

01

Identify the Target Reaction

The reaction to be solved for is the transformation of ethylene gas and chlorine gas into dichloroethane liquid, given by \(C_2H_4(g) + Cl_2(g) → C_2H_4Cl_2(l)\). This is the target reaction and we need to find \(\Delta H^{\circ}\) for it.

02

Rearrange and Combine the Known Reactions

The given reactions are: 1) \(4 HCl(g) + O_2(g) → 2 Cl_2(g) + 2 H_2O(l)\)with \(\Delta H^{\circ} = -202.4 kJ\)2) \(2 HCl(g) + C_2H_4(g) + 0.5 O_2(g) → C_2H_4Cl_2(l) + H_2O(l)\)with \(\Delta H^{\circ} = -318.7 kJ\)To match the target reaction, we need to manipulate the given reactions. Let's reverse the first reaction and halve it, which also reverses its enthalpy sign and halves the magnitude. Then, subtract the modified first reaction from the unmodified second reaction.

03

Solve for \(\Delta H^{\circ}\)

After the above manipulations, the resultant reaction is the target reaction and its \(\Delta H^{\circ}\) can be obtained by adding the enthalpy changes of the modified reactions, that is, half the reverse of the first reaction's \(\Delta H^{\circ}\) and the second reaction's \(\Delta H^{\circ}\).

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