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Chapter 13: Problem 27

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
For the given reactions at this temperature, the equilibrium constants are: a. \(K_a \approx 0.114\) b. \(K_b \approx 76.9\) c. \(K_c \approx 8.67\) d. \(K_d \approx 1.69 \times 10^{-4}\)

Step by step solution

01

Recall the basic properties of manipulating equilibrium constants

When manipulating a chemical reaction to create a new reaction, the equilibrium constant is also affected according to: 1. If the reaction is reversed, take the reciprocal of the equilibrium constant. 2. If the whole equation is multiplied or divided by a constant, raise the equilibrium constant to the power of that constant. We'll use these properties for calculating the values of \(K\) for each of the given reactions.
02

Calculate the equilibrium constant for reaction a

For reaction a, the initial reaction is divided by 2: $$\frac{1}{2} N_{2}(g) + \frac{3}{2} H_{2}(g) \rightleftharpoons NH_{3}(g)$$ Since the coefficients are multiplied by 1/2, we'll need to raise the initial equilibrium constant to the power of 1/2: $$K_a = \left(K\right)^\frac{1}{2} = \left(1.3 \times 10^{-2}\right)^\frac{1}{2} \approx 0.114$$
03

Calculate the equilibrium constant for reaction b

For reaction b, the initial reaction is reversed: $$2 NH_{3}(g) \rightleftharpoons N_{2}(g) + 3 H_{2}(g)$$ Since the reaction is reversed, take the reciprocal of the initial equilibrium constant: $$K_b = \frac{1}{K} = \frac{1}{1.3 \times 10^{-2}} \approx 76.9$$
04

Calculate the equilibrium constant for reaction c

For reaction c, the initial reaction is divided by 2 and then reversed: $$NH_{3}(g) \rightleftharpoons \frac{1}{2} N_{2}(g) + \frac{3}{2} H_{2}(g)$$ To find the equilibrium constant for this reaction, first raise the initial \(K\) to the power of 1/2, then take the reciprocal: $$K_c = \frac{1}{\left(K\right)^\frac{1}{2}} = \frac{1}{\left(1.3 \times 10^{-2}\right)^\frac{1}{2}} \approx 8.67$$
05

Calculate the equilibrium constant for reaction d

For reaction d, the initial reaction is multiplied by 2: $$2 N_{2}(g) + 6 H_{2}(g) \rightleftharpoons 4 NH_{3}(g)$$ Since the coefficients are multiplied by 2, raise the initial equilibrium constant to the power of 2: $$K_d = \left(K\right)^2 = \left(1.3 \times 10^{-2}\right)^2 \approx 1.69 \times 10^{-4}$$ Finally, we have the equilibrium constants for all the given reactions: a. \(K_a \approx 0.114\) b. \(K_b \approx 76.9\) c. \(K_c \approx 8.67\) d. \(K_d \approx 1.69 \times 10^{-4}\)

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Most popular questions from this chapter

The reaction to prepare methanol from carbon monoxide and hydrogen $$\mathrm{CO}(g)+\mathrm{H}_{2}(g) \leftrightharpoons \mathrm{CH}_{3} \mathrm{OH}(g)$$ is exothermic. If you wanted to use this reaction to produce methanol commercially, would high or low temperatures favor a maximum yield? Explain.

The equilibrium constant \(K_{\mathrm{p}}\) is \(2.4 \times 10^{3}\) at a certain temperature for the reaction $$2 \mathrm{NO}(g) \leftrightharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)$$ For which of the following sets of conditions is the system at equilibrium? For those not at equilibrium, in which direction will the system shift? a. $P_{\mathrm{NO}}=0.012 \mathrm{atm}, P_{\mathrm{N}_{2}}=0.11 \mathrm{atm}, P_{\mathrm{O}_{2}}=2.0 \mathrm{atm}$ b. $P_{\mathrm{NO}}=0.0078 \mathrm{atm}, P_{\mathrm{N}_{2}}=0.36 \mathrm{atm}, P_{\mathrm{O}_{2}}=0.67 \mathrm{atm}$ c. $P_{\mathrm{NO}}=0.0062 \mathrm{atm}, P_{\mathrm{N}_{2}}=0.51 \mathrm{atm}, P_{\mathrm{O}_{2}}=0.18 \mathrm{atm}$

Solid \(\mathrm{NH}_{4} \mathrm{HS}\) decomposes by the following endothermic process: $$\mathrm{NH}_{4} \mathrm{HS}(s) \leftrightharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g)$$ a. What effect will adding more \(\mathrm{NH}_{3}(g)\) have on the equilibrium? b. What effect will adding more \(\mathrm{NH}_{4} \mathrm{HS}(s)\) have on the equilibrium? c. What effect will increasing the volume of the container have on the equilibrium? d. What effect will decreasing the temperature have on the equilibrium?

Write the equilibrium expression (K) for each of the following gas-phase reactions. a. \(N_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)\) b. \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) c. $\operatorname{SiH}_{4}(g)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \operatorname{SiCl}_{4}(g)+2 \mathrm{H}_{2}(g)$ d. $2 \mathrm{PBr}_{3}(g)+3 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{PCl}_{3}(g)+3 \mathrm{Br}_{2}(g)$

Consider the following reaction: $$\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)$$ Amounts of \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CO}, \mathrm{H}_{2},\) and \(\mathrm{CO}_{2}\) are put into a flask so that the composition corresponds to an equilibrium position. If the CO placed in the flask is labeled with radioactive \(^{14} \mathrm{C}\) will \(^{14} \mathrm{C}\) be found only in \(\mathrm{CO}\) molecules for an indefinite period of time? Explain.
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