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Chapter 13: Problem 105

In a given experiment, 5.2 moles of pure NOCl were placed in an otherwise empty \(2.0-\mathrm{L}\) container. Equilibrium was established by the following reaction: $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \quad K=1.6 \times 10^{-5}$$ a. Using numerical values for the concentrations in the Initial row and expressions containing the variable \(x\) in both the Change and Equilibrium rows, complete the following table summarizing what happens as this reaction reaches equilibrium. Let \(x=\) the concentration of \(\mathrm{Cl}_{2}\) that is present at equilibrium. b. Calculate the equilibrium concentrations for all species.

Short Answer

Expert verified
The equilibrium concentrations for all species are: NOCl: \(2.5996 \thinspace M\) NO: \(0.0004 \thinspace M\) Cl₂: \(0.0002 \thinspace M\)

Step by step solution

01

Create an ICE table

To create an ICE table, the first step is to calculate the initial concentrations of each participant involved in the chemical reaction. For now, only NOCl has a given amount (5.2 moles), while the others are simply zero. Calculate NOCl's initial concentration: Initial Concentration of NOCl = Moles of NOCl / Volume of Container \(C_0(NOCl) = \frac {5.2 \thinspace mol }{2.0 \thinspace L}\) \(C_0(NOCl) = 2.6 \thinspace M\) Step 2: Complete the Initial Row of ICE Table
02

Fill out the ICE table for the initial concentrations

Now that we have the initial concentration of NOCl, we can fill out the initial row in the ICE table as follows: | | NOCl | 2 NO | Cl₂ | |---|------|------|------| | I | 2.6 | 0 | 0 | | C | | | | | E | | | | Step 3: Formulate the Change and Equilibrium Rows
03

Write the expressions for Change and Equilibrium rows

We know that, for every 2 moles of NOCl that react, we produce 2 moles of NO and 1 mole of Cl₂. So, when x moles of Cl₂ are formed, x moles of NO are formed and 2x moles of NOCl are consumed. Let's now write the expressions for the Change and Equilibrium rows: | | NOCl | 2 NO | Cl₂ | |---|---------|---------|-------| | I | 2.6 | 0 | 0 | | C | -2x | 2x | x | | E | 2.6-2x | 2x | x | Step 4: Write the Equilibrium Expression
04

Set up the equilibrium expression

Now that we have expressions for the equilibrium concentrations of all species, we will write the equilibrium expression using the equilibrium constant, K: \[K = \frac{[NO]^2 \times [Cl_2]}{[NOCl]^2}\] Plug in the equilibrium concentrations: \[1.6 \times 10^{-5} = \frac{(2x)^2 \times x}{(2.6-2x)^2}\] Step 5: Solve for x
05

Solve the equation to find x

By solving for x, we can find the concentration of Cl₂ at equilibrium, and also the equilibrium concentrations for the other species. Multiplying both sides by (2.6-2x)^2: \[(2.6-2x)^2= (2x)^2 \times x \times (1.6 \times 10^{-5})\] This is a quadratic equation in x, which we can solve to get: \(x \approx 0.0002 \thinspace M\) Step 6: Calculate Equilibrium Concentrations
06

Find the equilibrium concentrations

Using x, which is approximately 0.0002 M, we can now find the equilibrium concentrations of all species: \[NOCl = 2.6 - 2x \approx 2.6 - (2 \times 0.0002) \approx 2.5996 \thinspace M\] \[NO = 2x \approx 2 \times 0.0002 \approx 0.0004 \thinspace M\] \[Cl_2 = x \approx 0.0002 \thinspace M\] Thus, the equilibrium concentrations for all species are: NOCl: 2.5996 M NO: 0.0004 M Cl₂: 0.0002 M

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Most popular questions from this chapter

A sample of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is placed in an empty cylinder at \(25^{\circ} \mathrm{C}\) . After equilibrium is reached the total pressure is 1.5 atm and 16\(\%\) (by moles) of the original $\mathrm{N}_{2} \mathrm{O}_{4}(g)\( has dissociated to \)\mathrm{NO}_{2}(g) .$ a. Calculate the value of \(K_{\mathrm{p}}\) for this dissociation reaction at \(25^{\circ} \mathrm{C} .\) b. If the volume of the cylinder is increased until the total pressure is 1.0 atm (the temperature of the system remains constant), calculate the equilibrium pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(\mathrm{NO}_{2}(g) .\) c. What percentage (by moles) of the original $\mathrm{N}_{2} \mathrm{O}_{4}(g)$ is dissociated at the new equilibrium position (total pressure \(=1.00 \mathrm{atm} ) ?\)

At a given temperature, \(K=1.3 \times 10^{-2}\) for the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ Calculate values of \(K\) for the following reactions at this temperature. a. $\frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)$ b. $2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)$ c. $\mathrm{NH}_{3}(g) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)$ d. $2 \mathrm{N}_{2}(g)+6 \mathrm{H}_{2}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)$

For the reaction $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ \(K_{\mathrm{p}}=3.5 \times 10^{4}\) at 1495 \(\mathrm{K} .\) What is the value of \(K_{\mathrm{p}}\) for the following reactions at 1495 \(\mathrm{K}\) ? a. $\operatorname{HBr}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)$ b. \(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g)\) c. $\frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Br}_{2}(g) \rightleftharpoons \mathrm{HBr}(g)$

Consider the reaction $$\mathrm{CO}(g)+\mathrm{H}_{2}(g) \leftrightharpoons \mathrm{CH}_{3} \mathrm{OH}(g)$$ Suppose the system is at equilibrium, and then an additional mole of \(\mathrm{N}_{2} \mathrm{O}(g)\) is injected into the system at constant temperature. Once the reaction reestablishes equilibrium, has the amount of \(\mathrm{N}_{2} \mathrm{O}\) increased or decreased from its original equilibrium amount? Explain. What happens to the value of the equilibrium constant with this change?

The partial pressures of an equilibrium mixture of $\mathrm{N}_{2} \mathrm{O}_{4}(g)\( and \)\mathrm{NO}_{2}(g)\( are \)P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.34\( atm and \)P_{\mathrm{NO}_{2}}=1.20 \mathrm{atm}$ at a certain temperature. The volume of the container is doubled. Calculate the partial pressures of the two gases when a new equilibrium is established.
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