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Chapter 12: Problem 23

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. \(2 \mathrm{N}_{2}(g)+6 \mathrm{H}_{2}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)\).

Short Answer

Expert verified
The equilibrium constants for each reaction at the given temperature are: a. \(K_a = 0.114\) b. \(K_b = 76.9\) c. \(K_c = 8.74\) d. \(K_d = 1.69 \times 10^{-4}\)

Step by step solution

01

Relationship between equilibrium constants and stoichiometric coefficients

When multiplying a reaction by a constant, the equilibrium constant must be raised to the power of that same constant. For example, if we multiplied the given reaction by 2: \[ 2(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)) \] \[ (2N_2(g) + 6H_2(g) \rightleftharpoons 4NH_3(g)) \] Then the equilibrium constant would be: \[ K' = K^2 \] Now we can apply this to each of the reactions (a, b, c, d).
02

Calculate \(K\) for reaction a

In reaction a, each stoichiometric coefficient in the original reaction is multiplied by 0.5: \[ 0.5N_2(g) + 1.5H_2(g) \rightleftharpoons NH_3(g) \] The equilibrium constant for this reaction (\(K_a\)) will be: \[ K_a = K^{0.5} \] Here, we substitute the given K value and calculate \(K_a\): \[ K_a = (1.3 \times 10^{-2})^{0.5} \] \[ K_a = 0.114 \]
03

Calculate \(K\) for reaction b

In reaction b, the given reaction is simply reversed: \[ 2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g) \] The equilibrium constant for this reaction (\(K_b\)) will be the inverse of the original \(K\): \[ K_b = \frac{1}{K} \] Here, we substitute the given K value and calculate \(K_b\): \[ K_b = \frac{1}{1.3 \times 10^{-2}} \] \[ K_b = 76.9 \]
04

Calculate \(K\) for reaction c

In reaction c, the stoichiometric coefficients of the products of the original reaction are multiplied by 0.5, and the given reaction is reversed: \[ NH_3(g) \rightleftharpoons 0.5N_2(g) + 1.5H_2(g) \] This is equivalent to multiplying the original reaction by -0.5, so the equilibrium constant for this reaction (\(K_c\)) will be: \[ K_c = K^{-0.5} \] Here, we substitute the given K value and calculate \(K_c\): \[ K_c = (1.3 \times 10^{-2})^{-0.5} \] \[ K_c = 8.74 \]
05

Calculate \(K\) for reaction d

In reaction d, the given reaction is multiplied by 2: \[ 2N_2(g) + 6H_2(g) \rightleftharpoons 4NH_3(g) \] The equilibrium constant for this reaction (\(K_d\)) will be: \[ K_d = K^{2} \] Here, we substitute the given K value and calculate \(K_d\): \[ K_d = (1.3 \times 10^{-2})^{2} \] \[ K_d = 1.69 \times 10^{-4} \] Now we have the equilibrium constant values for each of the reactions (a, b, c, d) at the given temperature: a. \(K_a = 0.114\) b. \(K_b = 76.9\) c. \(K_c = 8.74\) d. \(K_d = 1.69 \times 10^{-4}\)

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Most popular questions from this chapter

At a particular temperature, 8.1 moles of \(\mathrm{NO}_{2}\) gas is placed in a 3.0 -L container. Over time the \(\mathrm{NO}_{2}\) decomposes to NO and \(\mathrm{O}_{2}:\) $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ At equilibrium the concentration of \(\mathrm{NO}(g)\) was found to be 1.4 mol/L. Calculate the value of \(K\) for this reaction.

For the following reaction at a certain temperature $$\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)$$.it is found that the equilibrium concentrations in a 5.00 -L rigid container are \(\left[\mathrm{H}_{2}\right]=0.0500\mathrm{M},\left[\mathrm{F}_{2}\right]=0.0100 \mathrm{M},\) and \([\mathrm{HF}]=\) \(0.400 \mathrm{M} .\) If 0.200 mole of \(\mathrm{F}_{2}\) is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished.

Consider the following exothermic reaction at equilibrium: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$. Predict how the following changes affect the number of moles of each component of the system after equilibrium is reestablished by completing the table below. Complete the table with the terms increase, decrease, or no change.

At a particular temperature, \(K=3.75\) for the reaction $$\mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g)$$.If all four gases had initial concentrations of 0.800 \(M,\) calculate the equilibrium concentrations of the gases.

Consider the following reactions:\(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g) \quad\) and \(\quad \mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow 2 \mathrm{HI}(g)\).List two property differences between these two reactions that relate to equilibrium.
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