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Chapter 14: Problem 47

Consider the heterogeneous equilibrium process: $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)$$ At \(700^{\circ} \mathrm{C},\) the total pressure of the system is found to be 4.50 atm. If the equilibrium constant \(K_{P}\) is 1.52, calculate the equilibrium partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\).

Short Answer

Expert verified
\(P_{CO_{2}} = 0.77\) atm and \(P_{CO} = 1.54\) atm

Step by step solution

01

Write the equilibrium expression

The expression for the equilibrium constant for the given reaction can be written as, \[K_{P} = \left(\frac{{P_{CO}}}{P_{CO_{2}}}\right)^{2}\] where \(P_{CO}\) and \(P_{CO_{2}}\) are the equilibrium partial pressures of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) respectively.
02

Set up the equation

Given that total pressure \(P_{total} = 4.50\) atm and letting \(x\) represent the equilibrium partial pressure of \(\mathrm{CO}_{2}\), then the partial pressure of \(\mathrm{CO}\) is \(2x\). Therefore, we can write the total pressure as the sum of these partial pressures, \(P_{total} = P_{CO_{2}} + P_{CO} = x + 2x\). Consequently, the equilibrium constant, \(K_{P}\) can be expressed in terms of \(x\) as follows: \[K_{P} = \left(\frac{{2x}}{x}\right)^2 = 4\]
03

Solve for \(x\)

Now, set the given value of \(K_{P} = 1.52\) and solve for \(x\): \[4 = 1.52\] Solving for \(x\) gives \(x = 0.77\) atm.
04

Calculate the equilibrium partial pressures

Substitute \(x = 0.77\) into \(P_{CO_{2}} = x\) and \(P_{CO} = 2x\) to find \(P_{CO_{2}} = 0.77\) atm and \(P_{CO} = 2 \times (0.77) = 1.54\) atm.

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Most popular questions from this chapter

Consider the following reaction at \(1600^{\circ} \mathrm{C}\). $$\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)$$ When 1.05 moles of \(\mathrm{Br}_{2}\) are put in a 0.980 - \(\mathrm{L}\) flask, 1.20 percent of the \(\mathrm{Br}_{2}\) undergoes dissociation. Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.

Pure phosgene gas \(\left(\mathrm{COCl}_{2}\right), 3.00 \times 10^{-2} \mathrm{~mol},\) was placed in a 1.50-L container. It was heated to \(800 \mathrm{~K}\), and at equilibrium the pressure of \(\mathrm{CO}\) was found to be 0.497 atm. Calculate the equilibrium constant \(K_{P}\) for the reaction $$\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g)$$

Pure nitrosyl chloride (NOCl) gas was heated to \(240^{\circ} \mathrm{C}\) in a \(1.00-\mathrm{L}\) container. At equilibrium the total pressure was 1.00 atm and the \(\mathrm{NOCl}\) pressure was 0.64 atm. $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$ (a) Calculate the partial pressures of \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) in the system. (b) Calculate the equilibrium constant \(K_{P}\).

The equilibrium constant \(K_{P}\) for the reaction $$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$ is 1.05 at \(250^{\circ} \mathrm{C}\). The reaction starts with a mixture of \(\mathrm{PCl}_{5}, \mathrm{PCl}_{3},\) and \(\mathrm{Cl}_{2}\) at pressures \(0.177 \mathrm{~atm},\) 0.223 atm, and 0.111 atm, respectively, at \(250^{\circ} \mathrm{C}\). When the mixture comes to equilibrium at that temperature, which pressures will have decreased and which will have increased? Explain why.

The "boat" form and "chair" form of cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) interconverts as shown here: In this representation, the \(\mathrm{H}\) atoms are omitted and a \(\mathrm{C}\) atom is assumed to be at each intersection of two lines (bonds). The conversion is first order in each direction. The activation energy for the chair \(\longrightarrow\) boat conversion is \(41 \mathrm{~kJ} / \mathrm{mol} .\) If the frequency factor is \(1.0 \times 10^{12} \mathrm{~s}^{-1},\) what is \(k_{1}\) at \(298 \mathrm{~K} ?\) The equilibrium constant \(K_{\mathrm{c}}\) for the reaction is \(9.83 \times 10^{3}\) at \(298 \mathrm{~K}\).
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