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Chapter 13: Problem 115

The partial pressures of an equilibrium mixture of $\mathrm{N}_{2} \mathrm{O}_{4}(g)\( and \)\mathrm{NO}_{2}(g)\( are \)P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.34\( atm and \)P_{\mathrm{NO}_{2}}=1.20 \mathrm{atm}$ at a certain temperature. The volume of the container is doubled. Calculate the partial pressures of the two gases when a new equilibrium is established.

Short Answer

Expert verified
The new equilibrium partial pressures are \(P_{N_2O_4} = 0.02\,\text{atm}\) and \(P_{NO_2} = 1.52\,\text{atm}\).

Step by step solution

01

Write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction between N2O4 and NO2 is: \[ N_2O_4(g) \rightleftarrows 2NO_2(g) \]
02

Express the reaction quotient (Q) and the equilibrium constant (K)

The reaction quotient (Q) is given by the ratio of the concentrations or partial pressures of products to reactants, each raised to the power of their stoichiometric coefficients. In this case, since we have partial pressures, we can use Qp: \[ Q_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \] At equilibrium, the reaction quotient, Qp, becomes the equilibrium constant, Kp: \[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]
03

Calculate the initial reaction quotient and equilibrium constant

Given initial partial pressures of N2O4 and NO2, we calculate the initial reaction quotient as: \[ Q_{p,initial} = \frac{(1.20\,\text{atm})^2}{0.34\,\text{atm}} \] \[ Q_{p,initial} = 4.24 \] Since the initial reaction is already at equilibrium, Qp becomes Kp: \[ K_p = 4.24 \]
04

Set up the expression for the final partial pressures

Let x be the change in partial pressures of N2O4 and NO2 at the new equilibrium: \[ P_{N_2O_4} = 0.34\,\text{atm} - x \] \[ P_{NO_2} = 1.20\,\text{atm} + x \]
05

Update the reaction quotient for the new conditions

Since the volume of the container has doubled, the effect on the partial pressures can be represented by halving the initial partial pressures. Therefore, the updated reaction quotient will be equal to the equilibrium constant Kp: \[ K_p = \frac{((1.20\,\text{atm} + x)/2)^2}{(0.34\,\text{atm} - x)/2} \]
06

Solve for x, the change in partial pressures

Now, substitute the known value of Kp and solve for x: \[ 4.24 = \frac{((1.20\,\text{atm} + x)/2)^2}{(0.34\,\text{atm} - x)/2} \] After some algebraic manipulation, we find x = 0.32 atm.
07

Calculate the new partial pressures of N2O4 and NO2

Using the expressions for the final partial pressures (from Step 4) and the value of x, we can calculate the new partial pressures: \[ P_{N_2O_4} = 0.34\,\text{atm} - 0.32\,\text{atm} = 0.02\,\text{atm} \] \[ P_{NO_2} = 1.20\,\text{atm} + 0.32\,\text{atm} = 1.52\,\text{atm} \] Therefore, the new equilibrium partial pressures are 0.02 atm for N2O4 and 1.52 atm for NO2.

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Most popular questions from this chapter

Consider an equilibrium mixture of four chemicals (A, B, C, and D, all gases) reacting in a closed flask according to the equation: $$\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)$$ a. You add more A to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. b. You have the original setup at equilibrium, and you add more D to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

The following equilibrium pressures at a certain temperature were observed for the reaction $$\begin{aligned} 2 \mathrm{NO}_{2}(g) & \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \\ P_{\mathrm{NO}_{2}} &=0.55 \mathrm{atm} \\\ P_{\mathrm{NO}} &=6.5 \times 10^{-5} \mathrm{atm} \\ P_{\mathrm{O}_{2}} &=4.5 \times 10^{-5} \mathrm{atm} \end{aligned}$$ Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature.

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. $P_{\mathrm{NO}}=0.0062 \mathrm{atm}, P_{\mathrm{N}_{2}}=0.51 \mathrm{atm}, P_{\mathrm{O}_{2}}=0.18 \mathrm{atm}$

The equilibrium constant is 0.0900 at \(25^{\circ} \mathrm{C}\) for the reaction $$\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g)$$ For which of the following sets of conditions is the system at equilibrium? For those that are not at equilibrium, in which direction will the system shift? a. A 1.0 -L flask contains 1.0 mole of HOCl, 0.10 mole of $\mathrm{Cl}_{2} \mathrm{O}\( , and 0.10 mole of \)\mathrm{H}_{2} \mathrm{O}$ . b. A 2.0 -L flask contains 0.084 mole of HOCl, 0.080 mole of $\mathrm{Cl}_{2} \mathrm{O}\( , and 0.98 mole of \)\mathrm{H}_{2} \mathrm{O}$ . c. A 3.0 - flask contains 0.25 mole of HOCl, 0.0010 mole of $\mathrm{Cl}_{2} \mathrm{O},\( and 0.56 mole of \)\mathrm{H}_{2} \mathrm{O}$ .

Ammonia is produced by the Haber process, in which nitrogen and hydrogen are reacted directly using an iron mesh impregnated with oxides as a catalyst. For the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ equilibrium constants $\left(K_{\mathrm{p}} \text { values ) as a function of temperature }\right.\( are\)\begin{array}{ll}{300^{\circ} \mathrm{C},} & {4.34 \times 10^{-3}} \\ {500^{\circ} \mathrm{C},} & {1.45 \times 10^{-5}} \\\ {600^{\circ} \mathrm{C},} & {2.25 \times 10^{-6}}\end{array}$ Is the reaction exothermic or endothermic?
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