To calculate the ΔH_rxn for the reaction:
[tex]\[ CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl(g), \][/tex]
we need to use the provided ΔH values for the given reactions:
1. [tex]\( \text{C(s) + 2H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)}, \, \Delta H = -74.6 \text{kJ} \)[/tex]
2. [tex]\( \text{C(s) + 2Cl}_2\text{(g)} \rightarrow \text{CCl}_4\text{(g)}, \, \Delta H = -95.7 \text{kJ} \)[/tex]
3. [tex]\( \text{H}_2\text{(g) + Cl}_2\text{(g)} \rightarrow 2 \text{HCl(g)}, \, \Delta H = -184.6 \text{kJ} \)[/tex]
We will use Hess's Law, which states that the enthalpy change of an overall reaction can be calculated by summing the enthalpy changes of individual steps.
The steps we need to obtain the target reaction are:
- Start from reaction 1 for the formation of CH_4(g).
- Add reaction 2 for the formation of CCl_4(g).
- Include four times reaction 3 for the formation of HCl(g) because the target reaction produces 4 moles of HCl.
Manipulate the reactions accordingly and calculate the overall ΔH:
1. [tex]\( \text{CH}_4\text{(g)} \)[/tex] formation:
[tex]\[ \text{C(s) + 2H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)}, \quad \Delta H_1 = -74.6 \text{kJ} \][/tex]
2. [tex]\( \text{CCl}_4\text{(g)} \)[/tex] formation:
[tex]\[ \text{C(s) + 2Cl}_2\text{(g)} \rightarrow \text{CCl}_4\text{(g)}, \quad \Delta H_2 = -95.7 \text{kJ} \][/tex]
3. Formation of 4 moles of HCl:
[tex]\[ 4 \cdot \left( \text{H}_2\text{(g) + Cl}_2\text{(g)} \rightarrow 2 \text{HCl(g)} \right), \quad \Delta H_3 = 4 \times (-184.6 \text{kJ}) = -738.4 \text{kJ} \][/tex]
Now, sum the enthalpy changes:
[tex]\[ \Delta H_{rxn} = \Delta H_2 + 4 \times \Delta H_3 - \Delta H_1 \][/tex]
[tex]\[ \Delta H_{rxn} = -95.7 \, \text{kJ} + (-738.4 \, \text{kJ}) - (-74.6 \, \text{kJ}) \][/tex]
Perform the arithmetic calculation:
[tex]\[ \Delta H_{rxn} = -95.7 \, \text{kJ} - 738.4 \, \text{kJ} + 74.6 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{rxn} = -95.7 - 738.4 + 74.6 \][/tex]
[tex]\[ \Delta H_{rxn} = -759.5 \, \text{kJ} \][/tex]
Thus, the enthalpy change for the reaction [tex]\( CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl(g) \)[/tex] is:
[tex]\[ \boxed{-759.5 \, \text{kJ}} \][/tex]
