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Chapter 13: Problem 102

An equilibrium mixture contains 0.60 g solid carbon and the gases carbon dioxide and carbon monoxide at partial pressures of 2.60 atm and 2.89 atm, respectively. Calculate the value of \(K_{\mathrm{p}}\) for the reaction $\mathrm{C}(s)+\mathrm{CO}_{2}(g) \Longrightarrow 2 \mathrm{CO}(g)$

Short Answer

Expert verified
The equilibrium constant (Kₚ) for the reaction C(s) + CO₂(g) ⇌ 2 CO(g) can be calculated by substituting the given partial pressures into the Kₚ expression: \(K_{\text{p}}=\frac{(\text{partial pressure of CO})^2}{(\text{partial pressure of CO}_2)}\). Plugging in the given values, we obtain \(K_{\mathrm{p}}=3.212\,\text{atm}\).

Step by step solution

01

Write the expression for Kₚ

For a given reaction at equilibrium, the expression for Kₚ is: \[ K_{\text{p}}=\frac{(\text{partial pressure of products})^{\mathrm{n}}}{(\text{partial pressure of reactants})^{\mathrm{m}}} \] where n and m are the stoichiometric coefficients of the products and reactants, respectively. For the given reaction: \[ \text{C}(s)+\text{CO}_2(g) \Longleftrightarrow 2 \text{CO}(g) \] The expression for Kₚ will be: \[ K_{\text{p}}=\frac{(\text{partial pressure of CO})^2}{(\text{partial pressure of CO}_2)} \]
02

Substitute the given values and solve for Kₚ

We are given the partial pressures of CO (2.89 atm) and CO₂ (2.60 atm). Let's plug these values into the expression for Kₚ: \[ K_{\text{p}}=\frac{(\text{2.89 atm})^2}{(\text{2.60 atm})} \] Now, we can calculate the value of Kₚ: \[ K_{\text{p}}=\frac{(2.89\,\text{atm})^2}{(2.60\,\text{atm})}= \frac{8.3521\,\text{atm}}{2.60\,\text{atm}} \] \[ K_{\mathrm{p}}=3.212\,\text{atm} \] The equilibrium constant, \(K_{\mathrm{p}}\), for the given reaction is \(3.212\,\text{atm}\).

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Most popular questions from this chapter

The reaction $$\mathrm{NH}_{4} \mathrm{SH}(s) \leftrightharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g)$$ has \(K_{\mathrm{p}}=0.10\) at \(27^{\circ} \mathrm{C} .\) What is the minimum amount of \(\mathrm{NH}_{4} \mathrm{SH}\) that must be present for this reaction to be at equilibrium in a 10.0 \(\mathrm{-L}\) container?

For the reaction $$2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)$$ \(K=2.4 \times 10^{-3}\) at a given temperature. At equilibrium in a \(2.0-\) L container it is found that $\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=1.1 \times 10^{-1} M\( and \)\left[\mathrm{H}_{2}(g)\right]=1.9 \times 10^{-2} \mathrm{M} .\( Calculate the moles of \)\mathrm{O}_{2}(g)$ present under these conditions.

An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature by the reaction $$3 \mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ At equilibrium, the concentrations are $\left[\mathrm{H}_{2}\right]=5.0 M,\left[\mathrm{N}_{2}\right]=$ \(8.0 M,\) and \(\left[\mathrm{NH}_{3}\right]=4.0 \mathrm{M} .\) What were the concentrations of nitrogen gas and hydrogen gas that were reacted initially?

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 moles each of \(\mathrm{NH}_{3}, \mathrm{H}_{2} \mathrm{S},\) and $\mathrm{NH}_{4} \mathrm{HS}\( are placed in a \)5.00-\mathrm{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?

For the reaction $\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g),$ consider two possibilities: (a) you mix 0.5 mole of each reactant, allow the system to come to equilibrium, and then add another mole of \(\mathrm{H}_{2}\) and allow the system to reach equilibrium again, or \((b)\) you \(\operatorname{mix} 1.5\) moles of \(\mathrm{H}_{2}\) and 0.5 mole of \(\mathrm{I}_{2}\) and allow the system to reach equilibrium. Will the final equilibrium mixture be different for the two procedures? Explain.
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