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Chapter 12: Problem 1

The following data are for the system $$\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)$$ $$\begin{array}{lcccccc}\hline \text { Time (s) } & 0 & 20 & 40 & 60 & 80 & 100 \\ P_{\mathrm{A}} \text { (atm) } & 1.00 & 0.83 & 0.72 & 0.65 & 0.62 & 0.62 \\ P_{B} \text { (atm) } & 0.00 & 0.34 & 0.56 & 0.70 & 0.76 & 0.76 \\\\\hline\end{array}$$ (a) How long does it take the system to reach equilibrium? (b) How does the rate of the forward reaction compare with the rate of the reverse reaction after 30 s? After 90 s?

Short Answer

Expert verified
Answer: The system reaches equilibrium at approximately 80 seconds. At 30 seconds, the forward reaction rate is higher than the reverse reaction rate. At 90 seconds, both the forward and reverse reaction rates are equal.

Step by step solution

01

Identify the equilibrium point

To find the equilibrium point, we need to look at the given data and find a moment when the partial pressures of A and B stop changing. From the data, we can see that this happens between 80 and 100 seconds, as the partial pressures of A and B remain constant at 0.62 atm and 0.76 atm, respectively.
02

Answer for (a)

It takes the system approximately 80 seconds to reach equilibrium.
03

Analyzing rates of forward and reverse reactions

To compare the rates of the forward and reverse reactions, we should look at the changes in partial pressures of A and B at the given times of 30 seconds and 90 seconds. At 30 seconds (which is between 20 and 40 seconds in the data), we can observe the partial pressures of A and B as follows: - Partial pressure of A: Decreases from 0.83 atm at 20 seconds to 0.72 atm at 40 seconds. - Partial pressure of B: Increases from 0.34 atm at 20 seconds to 0.56 atm at 40 seconds. At 90 seconds (which is between 80 and 100 seconds in the data), we can observe the partial pressures of A and B as follows: - Partial pressure of A: Remains constant at 0.62 atm. - Partial pressure of B: Remains constant at 0.76 atm.
04

Answer for (b)

After 30 seconds (between 20 and 40 seconds), the forward reaction rate (A converting to B) is still higher than the reverse reaction rate (B converting to A) since the partial pressure of A is decreasing and the partial pressure of B is increasing. After 90 seconds (between 80 and 100 seconds), both the forward and reverse reactions have the same rate, as the partial pressures of A and B have reached equilibrium and are not changing.

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Most popular questions from this chapter

At a certain temperature, nitrogen and oxygen gases combine to form nitrogen oxide gas. $$\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(g)$$ When equilibrium is established, the partial pressures of the gases are: \(P_{\mathrm{N}_{2}}=\) \(1.2 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.80 \mathrm{~atm}, P_{\mathrm{NO}}=0.022 \mathrm{~atm} .\) (a) Calculate \(K\) at the temperature of the reaction. (b) After equilibrium is reached, more oxygen is added to make its partial pressure \(1.2\) atm. Calculate the partial pressure of all gases when equilibrium is reestablished.

Given the following data at \(25^{\circ} \mathrm{C}\), $$\begin{array}{cl}2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & K=1 \times 10^{-30} \\ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) & K=8 \times 10^{1} \end{array}$$ calculate \(K\) for the formation of one mole of NOBr from its elements in the gaseous state at \(25^{\circ} \mathrm{C}\).

Solid ammonium iodide decomposes to ammonia and hydrogen iodide gases at sufficiently high temperatures. $$\mathrm{NH}_{4} \mathrm{I}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{HI}(g)$$ The equilibrium constant for the decomposition at \(673 \mathrm{~K}\) is \(0.215\). Fifteen grams of ammonium iodide are sealed in a \(5.0\) -L flask and heated to \(673 \mathrm{~K}\). (a) What is the total pressure in the flask at equilibrium? (b) How much ammonium iodide decomposes?

Given the following descriptions of reversible reactions, write a balanced equation (simplest whole-number coefficients) and the equilibrium constant expression \((K)\) for each. (a) Nitrogen gas reacts with solid sodium carbonate and solid carbon to produce carbon monoxide gas and solid sodium cyanide. (b) Solid magnesium nitride reacts with water vapor to form magnesium hydroxide solid and ammonia gas. (c) Ammonium ion in aqueous solution reacts with a strong base at \(25^{\circ} \mathrm{C}\), giving aqueous ammonia and water. (c) Hydrogen sulfide gas \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) bubbled into an aqueous solution of lead(II) ions produces lead sulfide precipitate and hydrogen ions.

Predict the direction in which each of the following equilibria will shift if the pressure on the system is decreased by expansion. (a) \(\mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g)\) (b) \(\mathrm{CI} \mathrm{F}_{5}(g) \rightleftharpoons \mathrm{Cl} \mathrm{F}_{3}(g)+\mathrm{F}_{2}(g)\) (c) \(\mathrm{HBr}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\)
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