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Chapter 13: Problem 32

At \(1100 \mathrm{~K}, K_{\mathrm{p}}=0.25\) for the reaction $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ What is the value of \(K\) at this temperature?

Short Answer

Expert verified
At 1100 K, the value of K for the given reaction is approximately \(K = 0.00002745\).

Step by step solution

01

Write the expression for the reaction quotient Qc, which is equal to K

Start by writing the expression for reaction quotient Qc in terms of concentrations for the given reaction: \( Q_c = \frac{[\mathrm{SO_3}]^2}{[\mathrm{SO_2}]^2 [\mathrm{O_2}]} \) At equilibrium, Qc = K, so \( K = \frac{[\mathrm{SO_3}]^2}{[\mathrm{SO_2}]^2 [\mathrm{O_2}]} \)
02

Write the expression for Kp and find its value

Kp can be written as: \( K_p = \frac{(\frac{P_{\mathrm{SO_3}}}{RT})^2}{(\frac{P_{\mathrm{SO_2}}}{RT})^2 (\frac{P_{\mathrm{O_2}}}{RT})} \) Given, Kp = 0.25. Now, we need to find the relationship between Kp and K.
03

Establish the relationship between Kp and K

To establish the relationship between Kp and K, we will divide both the expressions for K and Kp: \( \frac{K_p}{K} = \frac{(\frac{P_{\mathrm{SO_3}}^2}{(RT)^2})}{(\frac{P_{\mathrm{SO_2}}^2P_{\mathrm{O_2}}}{(RT)^3})} \) \( \frac{K_p}{K} = \frac{(\frac{P_{\mathrm{SO_3}}^2}{(RT)^3})}{(\frac{P_{\mathrm{SO_2}}^2P_{\mathrm{O_2}}}{(RT)^3})} \cdot \frac{RT}{RT} = \frac{P_{\mathrm{SO_2}}^2P_{\mathrm{O_2}}}{P_{\mathrm{SO_3}}^2} \cdot RT \)
04

Find the value of K

We know that Kp = 0.25, and from the reaction, the stoichiometric coefficients are equal so we will have delta n = 0, which means the ratio in the above expression will be 1. Thus, the relation becomes: \( \frac{0.25}{K} = RT\) Now, find the value of K at 1100 K. The temperature is given in Kelvin, so we will use the gas constant value R = 8.314 J/(mol路K): \( K = \frac{0.25}{(8.314 \,\mathrm{J/(mol\cdot K)})(1100 \,\mathrm{K})} \) Now, calculate the value of K: \( K = 0.00002745 \) So, at 1100 K, the value of K for the given reaction is approximately 0.00002745.

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Most popular questions from this chapter

The formation of peptide bonds is an important area of chemistry. The following reaction has an equilibrium constant \((K)\) of \(3.2 \times 10^{2}\) at some temperature: Alanine \((a q)+\) leucine \((a q) \rightleftharpoons\) alanine-leucine dipeptide \((a q)+\mathrm{H}_{2} \mathrm{O}(l)\) Which direction will this reaction need to shift to reach equilibrium under the following conditions? a. \([\) alanine \(]=0.60 M,[\) leucine \(]=0.40 M,[\) dipeptide \(]=0.20 M\) b. \([\) alanine \(]=3.5 \times 10^{-4} M,[\) leucine \(]=3.6 M,[\) dipeptide \(]=\) \(0.40 M\) c. \([\) alanine \(]=6.0 \times 10^{-3} M,[\) leucine \(]=9.0 \times 10^{-3} M\), \([\) dipeptide \(]=0.30 M\)

For the reaction $$ \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{HS}(s) $$ \(K=400\). at \(35.0^{\circ} \mathrm{C}\). If \(2.00 \mathrm{~mol}\) each of \(\mathrm{NH}_{3}, \mathrm{H}_{2} \mathrm{~S}\), and \(\mathrm{NH}_{4} \mathrm{HS}\) are placed in a 5.00-L vessel, what mass of \(\mathrm{NH}_{4} \mathrm{HS}\) will be present at equilibrium? What is the pressure of \(\mathrm{H}_{2} \mathrm{~S}\) at equilibrium?

Suppose \(K=4.5 \times 10^{-3}\) at a certain temperature for the reaction $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ If it is found that the concentration of \(\mathrm{PCl}_{5}\) is twice the concentration of \(\mathrm{PCl}_{3}\), what must be the concentration of \(\mathrm{Cl}_{2}\) under these conditions?

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.

At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}}=5.3 \times 10^{5}\) for the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ When a certain partial pressure of \(\mathrm{NH}_{3}(g)\) is put into an otherwise empty rigid vessel at \(25^{\circ} \mathrm{C}\), equilibrium is reached when \(50.0 \%\) of the original ammonia has decomposed. What was the original partial pressure of ammonia before any decomposition occurred?
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