Questions: Use the following information to answer the next question Some Equations C(s) + 2 H2(g) → CH4(g) ΔH = -74.8 kJ C(s) + 2 Cl2(g) → CCl4(l) ΔH = -106.4 kJ 1/2 H2(g) + 1/2 Cl2(g) → HCl(g) ΔH = -92.3 kJ The enthalpy change for the reaction CH4(g) + 4 Cl2(g) → CCl4(l) + 4 HCl(g) is -216.2 kJ -123.9 kJ -400.8 kJ -273.5 kJ

Use the following information to answer the next question Some Equations C(s) + 2 H2(g) → CH4(g) ΔH = -74.8 kJ C(s) + 2 Cl2(g) → CCl4(l) ΔH = -106.4 kJ 1/2 H2(g) + 1/2 Cl2(g) → HCl(g) ΔH = -92.3 kJ The enthalpy change for the reaction CH4(g) + 4 Cl2(g) → CCl4(l) + 4 HCl(g) is -216.2 kJ -123.9 kJ -400.8 kJ -273.5 kJ
Transcript text: Use the following information to answer the next question Some Equations \[ \begin{array}{ll} \mathrm{C}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CH}_{4}(\mathrm{~g}) & \Delta \mathrm{H}=-74.8 \mathrm{~kJ} \\ \mathrm{C}(\mathrm{s})+2 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CCl}_{4}(\mathrm{l}) & \Delta \mathrm{H}=-106.4 \mathrm{~kJ} \\ \frac{1}{2} \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{HCl}(\mathrm{g}) & \Delta \mathrm{H}=-92.3 \mathrm{~kJ} \end{array} \] 9. The enthalpy change for the reaction $\mathrm{CH}_{4}(\mathrm{~g})+4 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CCl}_{4}(\mathrm{I})+4 \mathrm{HCl}(\mathrm{g})$ is $-216.2 \mathrm{~kJ}$ $-123.9 \mathrm{~kJ}$ $-400.8 \mathrm{~kJ}$ $-273.5 \mathrm{~kJ}$
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Solution

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Solution Steps

Step 1: Identify the Target Reaction

The target reaction is:

\[ \mathrm{CH}_{4}(\mathrm{~g}) + 4 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CCl}_{4}(\mathrm{l}) + 4 \mathrm{HCl}(\mathrm{g}) \]

Step 2: Use Given Reactions to Formulate the Target Reaction

We need to manipulate the given reactions to derive the target reaction:

  1. \(\mathrm{C}(\mathrm{s}) + 2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CH}_{4}(\mathrm{~g})\) with \(\Delta \mathrm{H} = -74.8 \, \mathrm{kJ}\)
  2. \(\mathrm{C}(\mathrm{s}) + 2 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CCl}_{4}(\mathrm{l})\) with \(\Delta \mathrm{H} = -106.4 \, \mathrm{kJ}\)
  3. \(\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) + \frac{1}{2} \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{HCl}(\mathrm{g})\) with \(\Delta \mathrm{H} = -92.3 \, \mathrm{kJ}\)
Step 3: Reverse and Combine Reactions

  1. Reverse the first reaction to form \(\mathrm{CH}_{4}(\mathrm{~g})\):

    \[ \mathrm{CH}_{4}(\mathrm{~g}) \rightarrow \mathrm{C}(\mathrm{s}) + 2 \mathrm{H}_{2}(\mathrm{~g}) \quad \Delta \mathrm{H} = +74.8 \, \mathrm{kJ} \]

  2. Use the second reaction as is:

    \[ \mathrm{C}(\mathrm{s}) + 2 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CCl}_{4}(\mathrm{l}) \quad \Delta \mathrm{H} = -106.4 \, \mathrm{kJ} \]

  3. Multiply the third reaction by 4 to form 4 \(\mathrm{HCl}\):

    \[ 2 \mathrm{H}_{2}(\mathrm{~g}) + 2 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{HCl}(\mathrm{g}) \quad \Delta \mathrm{H} = 4 \times (-92.3) = -369.2 \, \mathrm{kJ} \]

Step 4: Add the Reactions

Add the modified reactions:

\[ \begin{align_} \mathrm{CH}_{4}(\mathrm{~g}) & \rightarrow \mathrm{C}(\mathrm{s}) + 2 \mathrm{H}_{2}(\mathrm{~g}) \quad (+74.8 \, \mathrm{kJ}) \\ \mathrm{C}(\mathrm{s}) + 2 \mathrm{Cl}_{2}(\mathrm{~g}) & \rightarrow \mathrm{CCl}_{4}(\mathrm{l}) \quad (-106.4 \, \mathrm{kJ}) \\ 2 \mathrm{H}_{2}(\mathrm{~g}) + 2 \mathrm{Cl}_{2}(\mathrm{~g}) & \rightarrow 4 \mathrm{HCl}(\mathrm{g}) \quad (-369.2 \, \mathrm{kJ}) \\ \hline \mathrm{CH}_{4}(\mathrm{~g}) + 4 \mathrm{Cl}_{2}(\mathrm{~g}) & \rightarrow \mathrm{CCl}_{4}(\mathrm{l}) + 4 \mathrm{HCl}(\mathrm{g}) \end{align_} \]

Step 5: Calculate the Enthalpy Change

Sum the enthalpy changes:

\[ \Delta \mathrm{H} = 74.8 - 106.4 - 369.2 = -400.8 \, \mathrm{kJ} \]

Final Answer

The enthalpy change for the reaction is \(\boxed{-400.8 \, \mathrm{kJ}}\).

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