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 $-216.2 \mathrm{~kJ}$ $-123.9 \mathrm{~kJ}$ $-400.8 \mathrm{~kJ}$ $-273.5 \mathrm{~kJ}$
failed

Solution

failed
failed

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 \(\boxed{-400.8 \, \mathrm{kJ}}\).

Was this solution helpful?
failed
Unhelpful
failed
Helpful

Questions asked by other users

failedAnswered
The estimated regression line and the standard error are given. Defects per Countertop = 6.717822 - 1.004950 (Hours of Training) sc = 1.229787 Suppose a new employee has had 10 hours of training, What would be the 99% prediction interval for the number of defects per countertop? Round your answer to two decimal places.
failedAnswered
Given the annual interest rate and the compounding period, find i, the interest rate per compounding period. 8.2% compounded quarterly i= % (Type an integer or decimal rounded to the nearest thousandth as needed.)
failedAnswered
Oxygen and nutrients are delivered to the outer portion of the retina by an extensive capillary network contained within the ora serrata choroid sclera ciliary body macula
failedAnswered
Keiana is determining if her message covers all the points in a logical and convincing order. Keiana is evaluating the of her message. A. tone B. scope C. organization D. content
failedAnswered
Problem 3 Thelma stands on top of a building whose height is H and throws a ball upward. The speed of the ball when it leaves her hand is vT, where the subscript stands for both Thelma and top. Bob is standing at the bottom of the building and at the same instant throws another ball upwards with speed vB. What is the slowest that vB can be if Bob's ball hits Thelma's when Thelma's ball is still on the way up? Express your answer in terms of gravitational acceleration g, H, and vT. (I would recommend that you draw a picture.)