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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 [H鈧俔_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 [H鈧俔_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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant (Kc)
The equilibrium constant, often denoted as \( K_c \), is a crucial concept in the study of chemical reactions. It quantifies the relationship between the concentrations of reactants and products at equilibrium. This allows chemists to predict the extent to which a reaction will proceed under given conditions.
For the reaction involving nitrogen monoxide (NO), hydrogen (H鈧�), nitrogen (N鈧�), and water vapor (H鈧侽), the equilibrium expression is formulated from the stoichiometry of the balanced chemical equation:\[ 2 \text{NO}(g) + 2 \text{H}_2(g) \rightleftharpoons \text{N}_2(g) + 2 \text{H}_2\text{O}(g) \]Here, the expression for \( K_c \) is:\[ K_c = \frac{[\text{N}_2][\text{H}_2\text{O}]^2}{[\text{NO}]^2[\text{H}_2]^2} \]In this equation, square brackets \([ ]\) indicate the molar concentrations of each species. The values reported must be at the equilibrium state. Interestingly, in this exercise, since H鈧� was entirely consumed, this equilibrium does not strictly exist, which is why the calculation of \( K_c \) becomes problematic.
Chemical Reactions
Chemical reactions are processes in which substances (reactants) convert into different substances (products). This transformation involves the making and breaking of chemical bonds. Reaction equations represent these processes and need to be balanced, meaning that the number of atoms for each element is the same on both sides of the equation.
In the exercise, the reaction of NO and H鈧� forms N鈧� and H鈧侽. This reversible reaction reaches a state called equilibrium, where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain stable.
Reactions can be influenced by several factors:
  • Concentration of reactants or products,
  • Temperature and pressure conditions, and
  • Presence of catalysts.
Understanding these aspects is essential for controlling industrial and laboratory processes, ensuring maximal yield and efficiency.
Stoichiometry
Stoichiometry refers to the calculation of reactants and products in chemical reactions. It is based on the conservation of mass and energy in chemical processes, making it a fundamental principle in chemistry.
In the reaction given in the exercise, stoichiometry is used to determine how much of each reactant is needed to produce a certain amount of product. It involves the coefficients found in the balanced chemical equation.For example, the reaction\[ 2 \text{NO}(g) + 2 \text{H}_2(g) \rightleftharpoons \text{N}_2(g) + 2 \text{H}_2\text{O}(g) \]shows that two moles of NO react with two moles of H鈧� to produce one mole of N鈧� and two moles of H鈧侽.
Using stoichiometry, we also deduced in the exercise that as NO decreases by \( 0.038 \mathrm{M} \), the corresponding decrease in H鈧� and increase in H鈧侽 and production of N鈧� could be calculated. Negative concentration outcomes imply a complete reaction, where one reactant is totally consumed. This demonstrates the limits and considerations of stoichiometric calculations in practice, especially when dealing with real-world systems.

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Most popular questions from this chapter

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) \(13 \mathrm{~atm} \mathrm{NH}_{2} 27 \mathrm{~atm} \mathrm{~N}_{2} 82 \mathrm{~atm} \mathrm{H}_{2}\)

At \(1000 \mathrm{~K}, K_{p}=1.85\) for the reaction $$ \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) $$ (a) What is the value of \(K_{p}\) for the reaction \(\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?(\mathbf{b})\) What is the value of \(K_{p}\) for the reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) ? (c) What is the value of \(K_{c}\) for the reaction in part (b)?

Consider the following equilibrium for which \(\Delta H<0\) $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ How will each of the following changes affect an equilibrium mixture of the three gases: (a) \(\mathrm{O}_{2}(g)\) is added to the system; (b) the reaction mixture is heated; (c) the volume of the reaction vessel is doubled; (d) a catalyst is added to the mixture; (e) the total pressure of the system is increased by adding a noble gas; (f) \(\mathrm{SO}_{3}(g)\) is removed from the system?

Suppose that the gas-phase reactions \(\mathrm{A} \longrightarrow \mathrm{B}\) and \(\mathrm{B} \longrightarrow \mathrm{A}\) are both elementary processes with rate constants of \(4.7 \times 10^{-3} \mathrm{~s}^{-1}\) and \(5.8 \times 10^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value of the equilibrium constant for the equilibrium \(\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g) ?\) (b) Which is greater at equilibrium, the partial pressure of \(\mathrm{A}\) or the partial pressure of \(\mathrm{B}\) ? Explain.

For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by either increasing the temperature or by increasing the volume of the reaction vessel. (a) Is the reaction exothermic or endothermic? (b) Does the balanced chemical equation have more molecules on the reactant side or product side?
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