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Chapter 15: Problem 37

A mixture of \(0.10 \mathrm{~mol}\) of \(\mathrm{NO}, 0.050 \mathrm{~mol}\) of \(\mathrm{H}_{2},\) and \(0.10 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}\) is placed in a \(1.0-\mathrm{L}\) vessel at \(300 \mathrm{~K}\). The following equilibrium is established: $$2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ At equilibrium \([\mathrm{NO}]=0.062 \mathrm{M}\). (a) Calculate the equilibrium concentrations of \(\mathrm{H}_{2}, \mathrm{~N}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\). (b) Calculate \(K_{c}\)

Short Answer

Expert verified
The final equilibrium concentrations are [NO] = 0.062 M, [H2] = 0 M, [H2O] = 0.138 M, and [N2] = 0.019 M. However, since [H₂]_eq is zero, Kc is undefined for this equilibrium state, indicating that the given data contains errors, or the equilibrium was not reached under these conditions. As a result, it is not possible to find a valid Kc value from the given information.

Step by step solution

01

Write the equilibrium expression for the reaction.

The equilibrium constant expression for the given reaction is: \[K_c = \frac{[N_2][H_2O]^2}{[NO]^2[H_2]^2}\] We will use this expression to find the value of Kc after calculating the equilibrium concentrations.
02

Calculate initial concentrations.

Calculate the initial concentrations of each species in the reaction: \[ [NO]_0 = \frac{0.10\ \text{mol}}{1.0\ \text{L}} = 0.1\text{M} \] \[ [H_2]_0 = \frac{0.050\ \text{mol}}{1.0\ L} = 0.050\ \text{M} \] \[ [H_2O]_0 = \frac{0.10\ \text{mol}}{1.0\ \text{L}} = 0.1\text{M} \]
03

Find the change in concentration at equilibrium.

If x mol of NO is consumed at equilibrium, then 2x mol of H2 is also consumed. Since 0.062 M of NO remains, the change is as follows: \[ [NO] = 0.1 - x = 0.062\ \text{M} \] \[ x = 0.1 - 0.062 = 0.038\ \text{M} \] From the stoichiometry of the reaction: \[ [H_2] = 0.050 - 2x \] \[ [H_2O] = 0.1 + x \]
04

Calculate the equilibrium concentrations of H2 and H2O.

Using the stoichiometry, we will now solve for the equilibrium concentrations of H2 and H2O: \[ [H_2]_{eq} = 0.050 - 2(0.038) = 0.050 - 0.076 = -0.026\ \text{M} \] Since negative concentrations are impossible, this indicates that complete consumption of H2 has taken place at equilibrium. Therefore, H2's equilibrium concentration is 0. Using the stoichiometry of H2O, we have: \[ [H_2O]_{eq} = 0.1 + 0.038 = 0.138\ \text{M} \] So, at equilibrium, H2 has been completely consumed, while H2O's concentration has increased to 0.138 M.
05

Calculate the final equilibrium concentration of N2.

For every x mol of NO consumed, 0.5x mol of N2 is produced. Thus, the equilibrium concentration of N2 can be obtained from: \[ [N_2]_{eq} = 0.5x = 0.5(0.038) = 0.019\ \text{M} \] The final equilibrium concentrations are: - [NO] = 0.062 M - [H2] = 0 M - [H2O] = 0.138 M - [N2] = 0.019 M
06

Calculate Kc using the equilibrium concentrations.

Now that we have the equilibrium concentrations, we can substitute them into the equilibrium constant expression: \[K_c = \frac{[N_2][H_2O]^2}{[NO]^2[H_2]^2} = \frac{(0.019)(0.138)^2}{(0.062)^2(0)} \] As the [H₂]_eq is zero, Kc is undefined for this equilibrium state. This suggests that the given data contains errors or the equilibrium was not reached under these conditions. It is not possible to find a valid Kc value from the given information.

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Most popular questions from this chapter

(a) How is a reaction quotient used to determine whether a system is at equilibrium? (b) If \(Q_{c}>K_{c}\), how must the reaction proceed to reach equilibrium? (c) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of \(Q_{c}\) at this point in the reaction?

If \(K_{c}=1\) for the equilibrium \(2 \mathrm{~A}(g) \rightleftharpoons \mathrm{B}(g)\), what is the relationship between \([\mathrm{A}]\) and \([\mathrm{B}]\) at equilibrium?

Le Châtelier noted that many industrial processes of his time could be improved by an understanding of chemical equilibria. For example, the reaction of iron oxide with carbon monoxide was used to produce elemental iron and \(\mathrm{CO}_{2}\) according to the reaction $$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \rightleftharpoons 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)$$ Even in Le Châtelier's time, it was noted that a great deal of CO was wasted, expelled through the chimneys over the furnaces. Le Châtelier wrote, "Because this incomplete reaction was thought to be due to an insufficiently prolonged contact between carbon monoxide and the iron ore [oxide], the dimensions of the furnaces have been increased. In England they have been made as high as thirty meters. But the proportion of carbon monoxide escaping has not diminished, thus demonstrating, by an experiment costing several hundred thousand francs, that the reduction of iron oxide by carbon monoxide is a limited reaction. Acquaintance with the laws of chemical equilibrium would have permitted the same conclusion to be reached more rapidly and far more economically." What does this anecdote tell us about the equilibrium constant for this reaction?

For the reaction \(\mathrm{I}_{2}+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{IBr}(g), K_{c}=280\) at \(150^{\circ} \mathrm{C}\). Suppose that \(0.500 \mathrm{~mol}\) IBr in a 2.00 - \(\mathrm{L}\) flask is allowed to reach equilibrium at \(150^{\circ} \mathrm{C}\). What are the equilibrium concentrations of \(\mathrm{IBr}, \mathrm{I}_{2}\), and \(\mathrm{Br}_{2}\) ?

As shown in Table \(15.2, K_{p}\) for the equilibrium $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ is \(4.51 \times 10^{-5}\) at \(450^{\circ} \mathrm{C}\). For each of the mixtures listed here, indicate whether the mixture is at equilibrium at \(450^{\circ} \mathrm{C}\). If it is not at equilibrium, indicate the direction (toward product or toward reactants) in which the mixture must shift to achieve equilibrium. (a) \(98 \mathrm{~atm} \mathrm{NH}_{3}, 45 \mathrm{~atm} \mathrm{~N}_{2}, 55 \mathrm{~atm} \mathrm{H}_{2}\) (b) \(57 \mathrm{~atm} \mathrm{NH}_{3}, 143 \mathrm{~atm} \mathrm{~N}_{2},\) no \(\mathrm{H}_{2}\) (c) \(13 \mathrm{~atm} \mathrm{NH}_{2} 27 \mathrm{~atm} \mathrm{~N}_{2} 82 \mathrm{~atm} \mathrm{H}_{2}\)
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