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Chapter 15: Problem 38

A flask is charged with 1.500 atm of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and 1.00 atm \(\mathrm{NO}_{2}(g)\) 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 0.512 atm. (a) What is the equilibrium partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (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 \(\mathrm{N}_{2}\mathrm{O}_{4}\) is 1.256 atm. The value of \(K_p\) for the reaction is 0.209, and the value of \(K_c\) for the reaction is 0.00850.

Step by step solution

01

Determine the change in pressure for both N2O4 and NO2

From the initial pressures, we know that the pressure of NO2 has increased from 1.00 atm to 0.512 atm. Since the stoichiometry of the reaction is 1:2 between N2O4 and NO2, for every mole of N2O4 that decomposes, two moles of NO2 are formed. Thus, we can find the change in pressure for both N2O4 and NO2 as follows: Change in pressure of NO2 = Final pressure of NO2 - Initial pressure of NO2 = 0.512 atm - 1.00 atm = -0.488 atm Change in pressure of N2O4 = 1/2 * Change in pressure of NO2 = 1/2 * (-0.488 atm) = -0.244 atm
02

Calculate the equilibrium partial pressure of N2O4

Now we can find the equilibrium partial pressure of N2O4 by adding the initial partial pressure and the change in pressure of N2O4: Equilibrium partial pressure of N2O4 = Initial pressure of N2O4 + Change in pressure of N2O4 = 1.500 atm - 0.244 atm = 1.256 atm
03

Calculate the value of Kp for the reaction

The equilibrium constant expression for the given reaction in terms of partial pressures (Kp) is: \[K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}}\] Using the equilibrium partial pressures calculated in Steps 1 and 2: \[K_p = \frac{(0.512 \, \text{atm})^2}{1.256 \, \text{atm}} = 0.209\] The value of Kp for the reaction is 0.209.
04

Calculate Kc for the reaction

To calculate Kc, we need to relate Kp to Kc using the ideal gas law: \[K_p = K_c(RT)^{\Delta n}\] where R is the gas constant (0.0821 L atm/mol K), T is the temperature in Kelvin (25°C = 298 K), and Δn is the change in the number of moles of gas (2 moles of NO2 - 1 mole of N2O4 = 1). Plugging in the values: \[0.209 = K_c(0.0821 \frac{\text{L atm}}{\text{mol K}})(298 \, K)^1\] Now, solve for Kc: \[K_c = \frac{0.209}{(0.0821)(298)} = 0.00850\] The value of Kc for the reaction is 0.00850.

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Most popular questions from this chapter

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be made by the reaction of \(\mathrm{CO}\) with \(\mathrm{H}_{2} :\) $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ (a) Use thermochemical data in Appendix C to calculate \(\Delta H^{\circ}\) for this reaction. (b) To maximize the equilibrium yield of methanol, would you use a high or low temperature? (c) To maximize the equilibrium yield of methanol, would you use a high or low pressure?

At \(100^{\circ} \mathrm{C}, K_{c}=0.078\) for the reaction $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{so}_{2}(g)+\mathrm{Cl}_{2}(g)$$ In an equilibrium mixture of the three gases, the concentrations of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) and \(\mathrm{SO}_{2}\) are 0.108 \(\mathrm{M}\) and \(0.052 \mathrm{M},\) respectively. What is the partial pressure of \(\mathrm{Cl}_{2}\) in the equilibrium mixture?

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?

At \(373 \mathrm{K}, K_{p}=0.416\) for the equilibrium $$2 \operatorname{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ If the pressures of \(\mathrm{NOBr}(g)\) and \(\mathrm{NO}(g)\) are equal, what is the equilibrium pressure of \(\mathrm{Br}_{2}(g) ?\)

Which of the following statements are true and which are false? (a) The equilibrium constant can never be a negative number. (b) In reactions that we draw with a single-headed arrow, the equilibrium constant has a value that is very close to zero. (c) As the value of the equilibrium constant increases, the speed at which a reaction reaches equilibrium increases.
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