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Chapter 15: Problem 38

A flask is charged with \(152.0 \mathrm{kPa}\) of $\mathrm{N}_{2} \mathrm{O}_{4}(g)\( and \)101.3 \mathrm{kPa}\( \)\mathrm{NO}_{2}(g)$ at \(25^{\circ} \mathrm{C},\) and the following equilibrium is achieved: $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ After equilibrium is reached, the partial pressure of \(\mathrm{NO}_{2}\) is \(51.9 \mathrm{kPa}\). (a) What is the equilibrium partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4} ?(\mathbf{b})\) Calculate the value of \(K_{p}\) for the reaction. (c) Calculate \(K_{c}\) for the reaction.

Short Answer

Expert verified
The equilibrium partial pressure of N2O4 is 176.7 kPa. The value of Kp for the reaction is approximately 15.19, and the value of Kc for the reaction is approximately 0.61.

Step by step solution

01

Write down the given initial partial pressures

The initial partial pressures of the gases are: N2O4(g): 152.0 kPa NO2(g): 101.3 kPa
02

Calculate the change in partial pressures at equilibrium

At equilibrium, the partial pressure of NO2 is 51.9 kPa. Therefore, the change in the partial pressure of NO2 is: ΔP_NO2 = final P_NO2 - initial P_NO2 = 51.9 kPa - 101.3 kPa = -49.4 kPa Since the stoichiometry of the reaction is 1 N2O4(g) to 2 NO2(g), the change in the partial pressure of N2O4 is half the change in the partial pressure of NO2: ΔP_N2O4 =1/2 * ΔP_NO2 = -49.4 kPa * (1/2) = 24.7 kPa
03

Calculate the equilibrium partial pressure of N2O4 and NO2

Equilibrium partial pressures: P_N2O4(eq) = initial P_N2O4 + ΔP_N2O4 = 152.0 kPa + 24.7 kPa = 176.7 kPa P_NO2(eq) = initial P_NO2 + ΔP_NO2 = 101.3 kPa - 49.4 kPa = 51.9 kPa
04

Calculate Kp for the reaction

Now that we have the equilibrium partial pressures, we can calculate the Kp of the reaction: Kp = (P_NO2(eq))^2 / P_N2O4(eq) = (51.9 kPa)² / 176.7 kPa ≈ 15.19
05

Calculate Kc for the reaction

To calculate Kc, we must relate Kp to Kc using the ideal gas law and the stoichiometry of the reaction: Kp = Kc * (RT)^(Δn) Where R is the ideal gas constant (R = 0.0821 L atm/mol K), T is the temperature in Kelvin, and Δn is the difference in moles of gaseous products and reactants in balanced chemical equation. For this reaction, T = 25°C = 298 K and Δn = 2 - 1 = 1 Kc = Kp / (RT)^(Δn) = 15.19 / ((0.0821 L atm/mol K) * (298 K))^(1) ≈ 0.61 Kc for the reaction is approximately 0.61 (rounded to two decimal places).

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Most popular questions from this chapter

For the equilibrium $$\mathrm{Br}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \operatorname{BrCl}(g) $$ at \(400 \mathrm{~K}, K_{c}=7.0 .\) If $0.25 \mathrm{~mol}\( of \)\mathrm{Br}_{2}\( and \)0.55 \mathrm{~mol}$ of \(\mathrm{Cl}_{2}\) are introduced into a 3.0-L container at \(400 \mathrm{~K}\), what will be the equilibrium concentrations of $\mathrm{Br}_{2}, \mathrm{Cl}_{2}\(, and \)\mathrm{BrCl}$ ?

The equilibrium constant for the reaction $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ is \(K_{c}=1.3 \times 10^{-2}\) at \(1000 \mathrm{~K}\). (a) At this temperature does the equilibrium favor \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\), or does it favor NOBr? (b) Calculate \(K_{c}\) for $2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) .$ (c) Calculate \(K_{c}\) for $\mathrm{NOBr}(g) \rightleftharpoons \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)$.

If \(K_{c}=1\) for the equilibrium $3 \mathrm{~A}(g) \rightleftharpoons 2 \mathrm{~B}(g)$, what is the relationship between [A] and [B] at equilibrium?

A mixture of \(1.374 \mathrm{~g}\) of \(\mathrm{H}_{2}\) and \(70.31 \mathrm{~g}\) of \(\mathrm{Br}_{2}\) is heated in a 2.00-L vessel at \(700 \mathrm{~K}\). These substances react according to $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ At equilibrium, the vessel is found to contain \(0.566 \mathrm{~g}\) of \(\mathrm{H}_{2}\). (a) Calculate the equilibrium concentrations of $\mathrm{H}_{2}, \mathrm{Br}_{2},\( and \)\mathrm{HBr} .$ (b) Calculate \(K_{c}\)

At \(25^{\circ} \mathrm{C}\), the reaction $$\mathrm{CaCrO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CrO}_{4}^{2-}(a q)$$ has an equilibrium constant \(K_{c}=7.1 \times 10^{-4}\). What are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and $\mathrm{CrO}_{4}{ }^{2-}\( in a saturated solution of \)\mathrm{CaCrO}_{4} ?$
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