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Chapter 15: Problem 74

At \(700 \mathrm{~K}\) the equilibrium constant for the reaction $$ \mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(\mathrm{s})+2 \mathrm{Cl}_{2}(g) $$ is \(K_{p}=0.76\). A flask is charged with \(2.00 \mathrm{~atm}\) of \(\mathrm{CCl}_{4}\), which then reaches equilibrium at \(700 \mathrm{~K}\). (a) What fraction of the \(\mathrm{CCl}_{4}\) is converted into \(\mathrm{C}\) and \(\mathrm{Cl}_{2} ?\) (b) What are the partial pressures of \(\mathrm{CCl}_{4}\) and \(\mathrm{Cl}_{2}\) at equilibrium?

Short Answer

Expert verified
(a) At equilibrium, \(12.5\%\) of the CCl4 is converted into C and Cl2. (b) The partial pressures at equilibrium are 1.75 atm for CCl4 and 0.50 atm for Cl2.

Step by step solution

01

Write the equilibrium expression

The equilibrium expression can be written using the equilibrium constant (Kp) which relates to the partial pressures of the species involved. For the reaction: \[CCl_4(g) \rightleftharpoons C(s) + 2 Cl_2(g)\] The equilibrium expression can be written as: \[K_p = \frac{P(Cl_2)^2}{P(CCl_4)}\]
02

Set up an ICE table

An ICE table helps us keep track of the initial, change and equilibrium partial pressures for each species. Using the given initial partial pressure of CCl4 (2.00 atm) and the stoichiometry of the reaction, we can set up the ICE table as follows: | Species | Initial | Change | Equilibrium | |:------------:|:---------:|:------------:|:-----------:| | CCl4 | 2.00 atm | -x | 2-x | | Cl2 | 0 | +2x | 2x | Where x represents the change in partial pressure that happens as the reaction proceeds and reaches equilibrium.
03

Substitute the values in the equilibrium expression

Now we can substitute the values from the ICE table into the equilibrium expression: \(0.76 = \frac{(2x)^2}{2-x}\)
04

Solve for x

To find x, we can solve the quadratic equation above: \(0.76(2-x) = 4x^2\) Expanding and simplifying: \(1.52 - 0.76x = 4x^2\) Rearranging to get a quadratic equation: \(4x^2 + 0.76x - 1.52 = 0\) Now, we can use the quadratic formula to solve for x: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Applying the formula with a = 4, b = 0.76, and c = -1.52: \(x = \frac{-0.76 \pm \sqrt{(0.76)^2 - 4(4)(-1.52)}}{2(4)}\) Solving for x, we get two possible values: x 鈮� 0.25 and x 鈮� -1.53. Since x represents the change in partial pressure, it cannot be negative. Therefore, we choose x 鈮� 0.25 as the correct value.
05

Answer the questions

(a) The fraction of CCl4 converted into C and Cl2 can be found by dividing the change (x) by the initial pressure of CCl4 (2.00 atm): Fraction converted = \( \frac{0.25}{2} = 0.125 \) Thus, 12.5% of the CCl4 is converted into C and Cl2. (b) To find the partial pressures at equilibrium, plug the value of x back into the equilibrium pressures from the ICE table: Partial pressure of CCl4 = 2 - x = 2 - 0.25 = 1.75 atm Partial pressure of Cl2 = 2x = 2(0.25) = 0.50 atm So, at equilibrium, the partial pressures are 1.75 atm for CCl4 and 0.50 atm for Cl2.

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Most popular questions from this chapter

(a) At \(1285{ }^{\circ} \mathrm{C}\) the equilibrium constant for the reaction \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\) is \(K_{c}=1.04 \times 10^{-3} .\) A \(0.200-\mathrm{L}\) vessel containing an equilibrium mixture of the gases has \(0.245 \mathrm{~g}\) \(\mathrm{Br}_{2}(g)\) in it. What is the mass of \(\mathrm{Br}(g)\) in the vessel? (b)For the reaction \(\mathrm{H}_{2}(g)+\mathrm{l}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g), K_{c}=55.3\) at \(700 \mathrm{~K}\). In a 2.00-L flask containing an equilibrium mixture of the three gases, there are \(0.056 \mathrm{~g} \mathrm{H}_{2}\) and \(4.36 \mathrm{~g} \mathrm{I}_{2}\). What is the mass of HI in the flask?

When \(1.50 \mathrm{~mol} \mathrm{CO}_{2}\) and \(1.50 \mathrm{~mol} \mathrm{H}_{2}\) are placed in a \(0.750-\mathrm{L}\) container at \(395^{\circ} \mathrm{C}\), the following equilibrium is achieved: \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) .\) If \(K_{c}=0.802\), what are the concentrations of each substance in the equilibrium mixture?

A flask is charged with \(1.500\) atm of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(1.00 \mathrm{~atm} \mathrm{NO}_{2}(g)\) at \(25^{\circ} \mathrm{C}\), and the following equilibrium is achieved: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ After equilibrium is reached, the partial pressure of \(\mathrm{NO}_{2}\) is \(0.512\) atm. (a) What is the equilibrium partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (b) Calculate the value of \(K_{p}\) for the reaction.

Consider the equilibrium \(\mathrm{IO}_{4}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}(a q), K_{c}=3.5 \times 10^{-2} .\) If you start with \(20.0 \mathrm{~mL}\) of a \(0.905 \mathrm{M}\) solution of \(\mathrm{NaIO}_{4}\), and then dilute it with water to \(250.0 \mathrm{~mL}\), what is the concentration of \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}\) at equilibrium?

An equilibrium mixture of \(\mathrm{H}_{2}, \mathrm{I}_{2}\), and \(\mathrm{HI}\) at \(458^{\circ} \mathrm{C}\) contains \(0.112 \mathrm{~mol} \mathrm{H}_{2}, 0.112 \mathrm{~mol} \mathrm{I}_{2}\), and \(0.775 \mathrm{~mol} \mathrm{HI}\) in a 5.00-L vessel. What are the equilibrium partial pressures when equilibrium is reestablished following the addition of \(0.100 \mathrm{~mol}\) of \(\mathrm{HI}\) ?
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