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Chapter 12: Problem 2

The following data are for the system $$\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)$$ $$\begin{array}{ccccccc}\hline \text { Time (s) } & 0 & 30 & 45 & 60 & 75 & 90 \\\ P_{\mathrm{A}} \text { (atm) } & 0.500 & 0.390 & 0.360 & 0.340 & 0.325 & 0.325 \\\ P_{\text {B }} \text { (atm) } & 0.000 & 0.220 & 0.280 & 0.320 & 0.350 & 0.350 \\\\\hline\end{array}$$ (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 45 s? After 90 s?

Short Answer

Expert verified
Answer: The system reached equilibrium at 75 seconds. At both 45 seconds and 90 seconds, the rates of the forward and reverse reactions were equal.

Step by step solution

01

(a) Identifying the equilibrium point

We need to find out when the pressure of A and B remains constant or changes very minimally. Let's examine the data: Time (s) - 0 | 30 | 45 | 60 | 75 | 90 P_A (atm) - 0.500 | 0.390 | 0.360 | 0.340 | 0.325 | 0.325 P_B (atm) - 0.000 | 0.220 | 0.280 | 0.320 | 0.350 | 0.350 From the data, we can observe that the equilibrium point is reached at 75 seconds as both A and B's pressures remained unchanged from 75 to 90 seconds.
02

(b) Rate comparison at 45 seconds

First, let's focus on the changes in pressure for A and B at 30 and 45 seconds. - Change in P_A: 0.390 atm to 0.360 atm - Change in P_B: 0.220 atm to 0.280 atm Since the stoichiometry of the reaction shows us that 1 mole of A produces 2 moles of B, for every decrease in pressure in A (0.030 atm), the pressure in B should increase by twice as much (0.060 atm). The observed change in P_B is actually 0.060 atm, which confirms that at 45 seconds, the forward and reverse reaction rates are equal.
03

(b) Rate comparison at 90 seconds

Now, let's consider the changes in pressure for A and B between 75 and 90 seconds. - Change in P_A: 0.325 atm to 0.325 atm - Change in P_B: 0.350 atm to 0.350 atm Since both P_A and P_B do not change from 75 to 90 seconds, this indicates that the system has reached equilibrium, and the forward and reverse reactions have the same rates. In summary, at both 45 seconds and 90 seconds, the forward and reverse reaction rates are equal.

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Most popular questions from this chapter

The reversible reaction between hydrogen chloride gas and one mole of oxygen gas produces steam and chlorine gas: $$4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=0.79$$ Predict the direction in which the system will move to reach equilibrium if one starts with (a) \(P_{\mathrm{H}_{2} \mathrm{O}}=P_{\mathrm{HCl}}=P_{\mathrm{O}_{2}}=0.20 \mathrm{~atm}\) (b) \(P_{\mathrm{HCl}}=0.30 \mathrm{~atm}, P_{\mathrm{H}_{2} \mathrm{O}}=0.35 \mathrm{~atm}, P_{\mathrm{Cl}_{2}}=0.2 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.15 \mathrm{~atm}\)

At a certain temperature, the reaction $$\mathrm{Xe}(g)+2 \mathrm{~F}_{2}(g) \rightleftharpoons \mathrm{XeF}_{4}(g)$$ gives a \(50.0 \%\) yield of \(\mathrm{XeF}_{4}\), starting with \(\mathrm{Xe}\left(P_{\mathrm{X}_{e}}=0.20 \mathrm{~atm}\right)\) and \(\mathrm{F}_{2}\) \(\left(P_{\mathrm{F}_{2}}=0.40 \mathrm{~atm}\right)\). Calculate \(K\) at this temperature. What must the initial pressure of \(\mathrm{F}_{2}\) be to convert \(75.0 \%\) of the xenon to \(\mathrm{XeF}_{4} ?\)

For the reaction $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ \(K\) at a certain temperature is \(0.50\). Predict the direction in which the system will move to reach equilibrium if one starts with (a) \(P_{\mathrm{O}_{2}}=P_{\mathrm{NO}}=P_{\mathrm{NO}_{2}}=0.10 \mathrm{~atm}\) (b) \(P_{\mathrm{NO}_{2}}=0.0848 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.0116 \mathrm{~atm}\) (c) \(P_{\mathrm{NO}_{2}}=0.20 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.010 \mathrm{~atm}, P_{\mathrm{NO}}=0.040 \mathrm{~atm}\)

Consider the equilibrium $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)$$ When this system is at equilibrium at \(700^{\circ} \mathrm{C}\) in a \(2.0\) - \(\mathrm{L}\) container, \(0.10 \mathrm{~mol}\) \(\mathrm{CO}, 0.20 \mathrm{~mol} \mathrm{CO}_{2}\), and \(0.40 \mathrm{~mol} \mathrm{C}\) are present. When the system is cooled to \(600^{\circ} \mathrm{C}\), an additional \(0.040 \mathrm{~mol} \mathrm{C}(s)\) forms. Calculate \(K\) at \(700^{\circ} \mathrm{C}\) and again at \(600^{\circ} \mathrm{C}\).

For the system $$\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)$$ at \(1000 \mathrm{~K}, K=0.45 .\) Sulfur trioxide, originally at \(1.00 \mathrm{~atm}\) pressure, partially dissociates to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) at \(1000 \mathrm{~K}\). What is its partial pressure at equilibrium?
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