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Chapter 13: Problem 29

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.

Short Answer

Expert verified
The equilibrium constant $K_p$ at this temperature is approximately \(6.29 \times 10^{-13}\).

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation is \(2 NO_2 (g) \rightleftharpoons 2 NO (g) + O_2 (g)\)
02

Write the expression for the equilibrium constant (Kp)

For the reaction, the expression for the equilibrium constant Kp is given by: \[K_p = \frac{(P_{NO})^2 \times P_{O_2}}{(P_{NO_2})^2}\]
03

Substitute the pressure values

Substitute the given pressure values in the Kp expression: \(P_{NO_2} = 0.55 \, atm\), \(P_{NO} = 6.5 \times 10^{-5} \, atm\), \(P_{O_2} = 4.5 \times 10^{-5} \, atm\) \[K_p = \frac{(6.5 \times 10^{-5})^2 \times (4.5 \times 10^{-5})}{(0.55)^2}\]
04

Calculate Kp

Perform the calculation to find the equilibrium constant Kp: \[K_p = \frac{(42.25 \times 10^{-10}) \times (4.5 \times 10^{-5})}{0.3025}\] \[K_p = \frac{190.125 \times 10^{-15}}{0.3025}\] \[K_p = 6.29 \times 10^{-13}\] The equilibrium constant Kp at this temperature is approximately \(6.29 \times 10^{-13}\).

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Most popular questions from this chapter

A sample of \(S_{8}(g)\) is placed in an otherwise empty rigid container at \(1325 \mathrm{~K}\) at an initial pressure of \(1.00 \mathrm{~atm}\), where it decomposes to \(\mathrm{S}_{2}(g)\) by the reaction $$ \mathrm{S}_{8}(g) \rightleftharpoons 4 \mathrm{~S}_{2}(g) $$ At equilibrium, the partial pressure of \(\mathrm{S}_{\mathrm{g}}\) is \(0.25 \mathrm{~atm} .\) Calculate \(K\). for this reaction at \(1325 \mathrm{~K}\).

Given the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)\), consider the following situations: i. You have \(1.3 M \mathrm{~A}\) and \(0.8 \mathrm{M}\) B initially. ii. You have \(1.3 M \mathrm{~A}, 0.8 \mathrm{M} \mathrm{B}\), and \(0.2 \mathrm{M} \mathrm{C}\) initially. iii. You have \(2.0 M \mathrm{~A}\) and \(0.8 M \mathrm{~B}\) initially. Order the preceding situations in terms of increasing equilibrium concentration of D. Explain your order. Then give the order in terms of increasing equilibrium concentration of \(\mathrm{B}\) and explain.

Consider the decomposition equilibrium for dinitrogen pentoxide: $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ At a certain temperature and a total pressure of \(1.00 \mathrm{~atm}\), the \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(0.50 \%\) decomposed (by moles) at equilibrium. a. If the volume is increased by a factor of \(10.0\), will the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposed at equilibrium be greater than, less than, or equal to \(0.50 \%\) ? Explain your answer. b. Calculate the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) that will be decomposed at equilibrium if the volume is increased by a factor of \(10.0 .\)

Le Châtelier's principle is stated (Section \(13.7)\) as follows: "If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change." The system \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) is used as an example in which the addition of nitrogen gas at equilibrium results in a decrease in \(\mathrm{H}_{2}\) concentration and an increase in \(\mathrm{NH}_{3}\) concentration. In the experiment the volume is assumed to be constant. On the other hand, if \(\mathrm{N}_{2}\) is added to the reaction system in a container with a piston so that the pressure can be held constant, the amount of \(\mathrm{NH}_{3}\) actually could decrease and the concentration of \(\mathrm{H}_{2}\) would increase as equilibrium is reestablished. Explain how this can happen. Also, if you consider this same system at equilibrium, the addition of an inert gas, holding the pressure constant, does affect the equilibrium position. Explain why the addition of an inert gas to this system in a rigid container does not affect the equilibrium position.

At \(900^{\circ} \mathrm{C}, K_{\mathrm{p}}=1.04\) for the reaction $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ At a low temperature, dry ice (solid \(\mathrm{CO}_{2}\) ), calcium oxide, and calcium carbonate are introduced into a \(50.0-\mathrm{L}\) reaction chamber. The temperature is raised to \(900^{\circ} \mathrm{C}\), resulting in the dry ice converting to gaseous \(\mathrm{CO}_{2}\). For the following mixtures, will the initial amount of calcium oxide increase, decrease, or remain the same as the system moves toward equilibrium at \(900^{\circ} \mathrm{C} ?\) a. \(655 \mathrm{~g} \mathrm{CaCO}_{3}, 95.0 \mathrm{~g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=2.55 \mathrm{~atm}\) b. \(780 \mathrm{~g} \mathrm{CaCO}_{3}, 1.00 \mathrm{~g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=1.04 \mathrm{~atm}\) c. \(0.14 \mathrm{~g} \mathrm{CaCO}_{3}, 5000 \mathrm{~g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=1.04 \mathrm{~atm}\) d. \(715 \mathrm{~g} \mathrm{CaCO}_{3}, 813 \mathrm{~g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=0.211 \mathrm{~atm}\)
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