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Chapter 13: Problem 43

In a study of the reaction $$3 \mathrm{Fe}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+4 \mathrm{H}_{2}(g)$$ at 1200 \(\mathrm{K}\) it was observed that when the equilibrium partial pressure of water vapor is 15.0 torr, the total pressure at equilibrium is 36.3 torr. Calculate the value of \(K_{\mathrm{p}}\) for this reaction at 1200 \(\mathrm{K}\) . Hint: Apply Dalton's law of partial pressures.)

Short Answer

Expert verified
The value of \(K_p\) for the given reaction at 1200 K is 2.19.

Step by step solution

01

Write down the Kp expression for the given reaction

According to the balanced equation, $$ K_\mathrm{p} = \frac{(\mathrm{P_{H_2}})^4}{(\mathrm{P_{H_2O}})^4} $$
02

Calculate the Partial Pressure of H2

We have the total pressure at equilibrium, P_total = 36.3 torr and the equilibrium partial pressure of water vapor, P_H2O = 15.0 torr. According to Dalton's law, P_total = P_Fe3O4 + P_H2 + P_H2O. Since Fe3O4 is a solid, it will not contribute to the total pressure, so P_total = P_H2 + P_H2O. Solving for P_H2: $$ \mathrm{P_{H_2}} = \mathrm{P_{total}} - \mathrm{P_{H_2O} }= 36.3 \mathrm{~torr} - 15.0 \mathrm{~torr} = 21.3 \mathrm{~torr} $$
03

Calculate the Kp

Substitute the values of P_H2 and P_H2O obtained in the previous steps into the Kp expression: $$ K_\mathrm{p} = \frac{(\mathrm{P_{H_2}})^4}{(\mathrm{P_{H_2O}})^4} = \frac{(21.3 \ \mathrm{torr})^4}{(15.0 \ \mathrm{torr})^4} = 2.19 $$ The value of Kp for this reaction at 1200 K is 2.19.

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Most popular questions from this chapter

At a particular temperature, \(K=2.0 \times 10^{-6}\) for the reaction $$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$$ If 2.0 moles of \(\mathrm{CO}_{2}\) is initially placed into a 5.0 -L vessel, calculate the equilibrium concentrations of all species.

Explain why the development of a vapor pressure above a liquid in a closed container represents an equilibrium. What are the opposing processes? How do we recognize when the system has reached a state of equilibrium?

The equilibrium constant is 0.0900 at \(25^{\circ} \mathrm{C}\) for the reaction $$\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g)$$ For which of the following sets of conditions is the system at equilibrium? For those that are not at equilibrium, in which direction will the system shift? a. A 1.0 -L flask contains 1.0 mole of HOCl, 0.10 mole of $\mathrm{Cl}_{2} \mathrm{O}\( , and 0.10 mole of \)\mathrm{H}_{2} \mathrm{O}$ . b. A 2.0 -L flask contains 0.084 mole of HOCl, 0.080 mole of $\mathrm{Cl}_{2} \mathrm{O}\( , and 0.98 mole of \)\mathrm{H}_{2} \mathrm{O}$ . c. A 3.0 - flask contains 0.25 mole of HOCl, 0.0010 mole of $\mathrm{Cl}_{2} \mathrm{O},\( and 0.56 mole of \)\mathrm{H}_{2} \mathrm{O}$ .

Consider the decomposition equilibrium for dinitrogen pentoxide: $$2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$$ At a certain temperature and a total pressure of 1.00 atm, the $\mathrm{N}_{2} \mathrm{O}_{5}\( is 0.50\)\%$ decomposed (by moles) at equilibrium. a. If the volume is increased by a factor of \(10.0,\) will the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposed at equilibrium be greater than, less than, or equal to 0.50\(\% ?\) Explain your answer. b. Calculate the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) that will be decomposed at equilibrium if the volume is increased by a factor of \(10.0 .\)

At a particular temperature, 8.0 moles of \(\mathrm{NO}_{2}\) is placed into a 1.0 -L container and the \(\mathrm{NO}_{2}\) dissociates by the reaction $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ At equilibrium the concentration of \(\mathrm{NO}(g)\) is 2.0 \(\mathrm{M}\) . Calculate \(K\) for this reaction.
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