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Chapter 13: Problem 110

The equilibrium constant \(K_{\mathrm{p}}\) for the reaction $$ \mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g) $$ at \(700^{\circ} \mathrm{C}\) is \(0.76\) atm. Determine the initial pressure of carbon tetrachloride that will produce a total equilibrium pressure of \(1.20 \mathrm{~atm}\) at \(700^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The initial pressure of carbon tetrachloride that will produce a total equilibrium pressure of 1.20 atm at 700°C is approximately \( 1.027 \, atm \).

Step by step solution

01

Write the balanced chemical equation and the expression for Kp

Here is the balanced chemical equation and the Kp expression: \( CCl_4(g) \rightleftharpoons C(s) + 2Cl_2(g) \) \[ K_p = \frac{[Cl_2]^2}{[CCl_4]} \]
02

Setup the ICE table

To set up the ICE table, we can represent the initial pressure of CCl4 as P and all the other components as 0 since we don't have any information about them. | | CCl4 | C | Cl2 | |----------|-------|---------|--------| | Initial | P | 0 | 0 | | Change | -x | +x | +2x | | Equilbr. | P - x | x | 2x | Since carbon solid is not involved in the Kp expression, it won't affect the table.
03

Substitute the equilibrium values from the ICE table into the Kp expression

Substitute the equilibrium values from the ICE table into the Kp expression: \[ K_p = \frac{(2x)^2}{(P-x)} \]
04

Use the given Kp value to solve for x

Since Kp = 0.76 at 700°C, we substitute this into the equation from Step 3 and solve for x: \[ 0.76 = \frac{(2x)^2}{(P-x)} \]
05

Solve for the total pressure at equilibrium

The total equilibrium pressure is given as 1.20 atm. This includes the pressure of the CCl4 and the Cl2 at equilibrium: \[ P_{total} = (P-x) + 2x = 1.20 \]
06

Solve the system of equations to find the initial pressure of CCl4

We have a system of two equations with two variables (P and x): 1) \( 0.76 = \frac{(2x)^2}{(P-x)} \) 2) \( 1.20 = (P-x) + 2x \) First, solve Eq. (2) for P: \( P = 1.20 - x \) Then, substitute this into Eq. (1) and solve for x: \[ 0.76 = \frac{(2x)^2}{(1.20 - x) - x} \] Solve for x, approximately 0.173 atm and substitute this value back into the equation for P above: \( P = 1.20 - 0.173 \)
07

Calculate the initial pressure of CCl4

Finally, we can calculate the initial pressure of CCl4: \( P = 1.027 \, atm \) Thus, the initial pressure of carbon tetrachloride that will produce a total equilibrium pressure of 1.20 atm at 700°C is approximately 1.027 atm.

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Most popular questions from this chapter

Suppose \(K=4.5 \times 10^{-3}\) at a certain temperature for the reaction $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ If it is found that the concentration of \(\mathrm{PCl}_{5}\) is twice the concentration of \(\mathrm{PCl}_{3}\), what must be the concentration of \(\mathrm{Cl}_{2}\) under these conditions?

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The formation of peptide bonds is an important area of chemistry. The following reaction has an equilibrium constant \((K)\) of \(3.2 \times 10^{2}\) at some temperature: Alanine \((a q)+\) leucine \((a q) \rightleftharpoons\) alanine-leucine dipeptide \((a q)+\mathrm{H}_{2} \mathrm{O}(l)\) Which direction will this reaction need to shift to reach equilibrium under the following conditions? a. \([\) alanine \(]=0.60 M,[\) leucine \(]=0.40 M,[\) dipeptide \(]=0.20 M\) b. \([\) alanine \(]=3.5 \times 10^{-4} M,[\) leucine \(]=3.6 M,[\) dipeptide \(]=\) \(0.40 M\) c. \([\) alanine \(]=6.0 \times 10^{-3} M,[\) leucine \(]=9.0 \times 10^{-3} M\), \([\) dipeptide \(]=0.30 M\)

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