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Chapter 15: Problem 33

The equilibrium \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\) is established at 500 \(\mathrm{K}\) . An equilibrium mixture of the three gases has partial pressures of 0.095 atm, 0.171 atm, and 0.28 atm for \(\mathrm{NO}, \mathrm{Cl}_{2},\) and \(\mathrm{NOCl}\) , respectively. (a) Calculate \(K_{p}\) for this reaction at 500.0 \(\mathrm{K}\) . (b) If the vessel has a volume of 5.00 \(\mathrm{L}\) , calculate \(K_{c}\) at this temperature.

Short Answer

Expert verified
a) The equilibrium constant, \(K_p\), at 500.0 K is 9.128. b) The equilibrium constant, \(K_c\), at 500.0 K is 35.01.

Step by step solution

01

Write the expression for Kp

We can write the expression for Kp of the given reaction as: \[K_p = \frac{P_{NOCl}^2}{P_{NO}^2 \times P_{Cl_2}}\] where \(P_{NOCl}, P_{NO},\text{ and } P_{Cl_2}\) are the partial pressures of the corresponding gases at equilibrium.
02

Calculate Kp using given pressures

Now we can use the given partial pressures at equilibrium to calculate Kp at 500 K: \[K_p = \frac{(0.28)^2}{(0.095)^2 \times 0.171} \] Using a calculator, we find that: \[K_p = 9.128\]
03

Calculate moles of each species at equilibrium

We are given the volume of the vessel is 5.00 L, and the equilibrium partial pressures. To find the moles of each species at equilibrium, we can use the Ideal Gas Law formula \(PV = nRT\), rearranged as: \[n = \frac{PV}{RT}\] Given that the temperature is 500.0 K and the ideal gas constant is R = 0.0821 atm L mol\(^{-1}\) K\(^{-1}\), we can find the number of moles for each species at equilibrium: \[n_{NO} = \frac{(0.095)(5.00)}{(0.0821)(500)} \Rightarrow n_{NO} = 0.0115\text{ mol}\] \[n_{Cl_2} = \frac{(0.171)(5.00)}{(0.0821)(500)} \Rightarrow n_{Cl_2} = 0.0208\text{ mol}\] \[n_{NOCl} = \frac{(0.28)(5.00)}{(0.0821)(500)} \Rightarrow n_{NOCl} = 0.0340\text{ mol}\]
04

Write the expression for Kc

Now we will write the equilibrium constant, Kc, expression for the given reaction: \[K_c = \frac{[NOCl]^2}{[NO]^2 \times [Cl_2]}\] where \([NOCl], [NO],\text{ and } [Cl_2]\) represent the concentration (in mol L\(^{-1}\)) of the corresponding gases at equilibrium.
05

Calculate Kc using the equilibrium moles

With the equilibrium moles and the volume of 5.00 L, we can calculate the equilibrium concentrations: \[[NO] = \frac{n_{NO}}{V} = \frac{0.0115}{5.00} = 0.0023\text{ mol L}^{-1}\] \[[Cl_2] = \frac{n_{Cl_2}}{V} = \frac{0.0208}{5.00} = 0.00416\text{ mol L}^{-1}\] \[[NOCl] = \frac{n_{NOCl}}{V} = \frac{0.0340}{5.00} = 0.0068\text{ mol L}^{-1}\] Now, we can plug these values into the Kc expression: \[K_c = \frac{(0.0068)^2}{(0.0023)^2 \times 0.00416}\] Using a calculator, we find that: \[K_c = 35.01\] To summarize: a) The equilibrium constant, Kp, at 500.0 K is 9.128. b) The equilibrium constant, Kc, at 500.0 K is 35.01.

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Most popular questions from this chapter

Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps: $$\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \operatorname{COCl}(g)+\mathrm{Cl}(g)$$ At \(25^{\circ} \mathrm{C},\) the rate constants for the forward and reverse reactions are \(1.4 \times 10^{-28} M^{-1} \mathrm{s}^{-1}\) and \(9.3 \times 10^{10} M^{-1} \mathrm{s}^{-1}\) respectively. (a) What is the value for the equilibrium constant at \(25^{\circ} \mathrm{C} ?\) (b) Are reactants or products more plentiful at equilibrium?

The reaction of an organic acid with an alcohol, in organic solvent, to produce an ester and water is commonly done in the pharmaceutical industry. This reaction is catalyzed by strong acid (usually \(\mathrm{H}_{2} \mathrm{SO}_{4} ) .\) A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water: $$\begin{aligned} \mathrm{CH}_{3} \mathrm{COOH}(s o l v)+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(s o l v) & \rightleftharpoons \\ \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(\mathrm{solv}) &+\mathrm{H}_{2} \mathrm{O}(\text {solv}) \end{aligned}$$ where \(^{a}(s o l v)^{\prime \prime}\) indicates that all reactants and products are in solution but not an aqueous solution. The equilibrium constant for this reaction at \(55^{\circ} \mathrm{C}\) is 6.68 . A pharmaceutical chemist makes up 15.0 \(\mathrm{L}\) of a solution that is initially 0.275 \(\mathrm{M}\) in acetic acid and 3.85\(M\) in ethanol. At equilibrium, how many grams of ethyl acetate are formed?

A mixture of 0.2000 mol of \(\mathrm{CO}_{2}, 0.1000 \mathrm{mol}\) of \(\mathrm{H}_{2},\) and 0.1600 mol of \(\mathrm{H}_{2} \mathrm{O}\) is placed in a 2.000 -L vessel. The following equilibrium is established at \(500 \mathrm{K} :\) $$\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ (a) Calculate the initial partial pressures of \(\mathrm{CO}_{2}, \mathrm{H}_{2,}\) and \(\mathrm{H}_{2} \mathrm{O}\) (b) At equilibrium \(P_{\mathrm{H}_{2} \mathrm{O}}\) \(=3.51\) atm. Calculate the equilibrium partial pressures of \(\mathrm{CO}_{2}, \mathrm{H}_{2},\) and \(\mathrm{CO} .(\mathbf{c})\) Calculate \(K_{p}\) for the reaction. (d) Calculate \(K_{c}\) for the reaction.

For the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g), K_{c}=55.3\) at 700 \(\mathrm{K} .\) In a \(2.00-\mathrm{L}\) flask containing an equilibrium mixture of the three gases, there are 0.056 \(\mathrm{g} \mathrm{H}_{2}\) and 4.36 \(\mathrm{g} \mathrm{I}_{2}\) . What is the mass of HI in the flask?

The equilibrium constant for the dissociation of molecular iodine, \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g),\) at 800 \(\mathrm{K}\) is \(K_{c}=3.1 \times 10^{-5} .\) (a) Which species predominates at equilibrium \(\mathrm{I}_{2}\) or \(\mathrm{I}\) ? (b) Assuming both forward and reverse reactions are elementary processes, which reaction has the larger rate constant, the forward or the reverse reaction?
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