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Chapter 13: Problem 74

Predict the shift in the equilibrium position that will occur for each of the following reactions when the volume of the reaction container is increased. a. $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$ b. $\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$ c. \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\) d. \(\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)\) e. $\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$

Short Answer

Expert verified
a. The equilibrium position will shift to the left. b. The equilibrium position will shift to the right. c. There will be no shift in the equilibrium position. d. The equilibrium position will shift to the right. e. The equilibrium position will shift to the right.

Step by step solution

01

Reaction a: Analyzing the equilibrium condition

For the equilibrium reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\), notice that there are 4 moles of gas on the left-hand side (1 mole of \(\mathrm{N}_{2}\) and 3 moles of \(\mathrm{H}_{2}\)) and 2 moles of \(\mathrm{NH}_{3}\) gas on the right-hand side, for a total of 4 moles on the left and 2 moles on the right.
02

Reaction a: Predicting the shift

When the volume of the reaction container is increased, the pressure decreases. According to Le Chatelier's principle, the equilibrium will shift in the direction that counteracts this pressure change. In this case, the reaction will shift in the direction that has more moles of gas, which is the left-hand side (the reactant side). Therefore, the equilibrium position will shift to the left.
03

Reaction b: Analyzing the equilibrium condition

For the equilibrium reaction \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\), there is 1 mole of \(\mathrm{PCl}_{5}\) gas on the left-hand side and 1 mole of \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) each on the right-hand side (2 moles in total).
04

Reaction b: Predicting the shift

Upon increasing the volume of the reaction container and decreasing the pressure, the equilibrium will shift in the direction with more moles of gas molecules, which is the right-hand side (the product side) in this case. Therefore, the equilibrium position will shift to the right.
05

Reaction c: Analyzing the equilibrium condition

For the equilibrium reaction \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\), there are 2 moles of gas on both the left-hand and right-hand sides (1 mole each of \(\mathrm{H}_{2}\) and \(\mathrm{F}_{2}\) on the left, and 2 moles of \(\mathrm{HF}\) on the right).
06

Reaction c: Predicting the shift

Since there are equal moles of gas molecules on both sides of the equilibrium, there will be no shift in the equilibrium position when the volume of the reaction container is increased.
07

Reaction d: Analyzing the equilibrium condition

For the equilibrium reaction \(\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)\), there is 1 mole of \(\mathrm{COCl}_{2}\) gas on the left-hand side and 1 mole of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) each on the right-hand side (2 moles in total).
08

Reaction d: Predicting the shift

Upon increasing the volume of the reaction container and decreasing the pressure, the equilibrium will shift to the side with more moles of gas molecules, which is the right-hand side (the product side), in this case. Therefore, the equilibrium position will shift to the right.
09

Reaction e: Analyzing the equilibrium condition

For the equilibrium reaction \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\), there are no moles of gas on the left-hand side and 1 mole of \(\mathrm{CO}_{2}\) gas on the right-hand side.
10

Reaction e: Predicting the shift

When the volume of the reaction container is increased, the pressure decreases. The equilibrium will shift to the side with more moles of gas molecules, which is the right-hand side (the product side) in this case. Therefore, the equilibrium position will shift to the right.

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Most popular questions from this chapter

Which of the following statements is(are) true? Correct the false statement(s). a. When a reactant is added to a system at equilibrium at a given temperature, the reaction will shift right to reestablish equilibrium. b. When a product is added to a system at equilibrium at a given temperature, the value of K for the reaction will increase when equilibrium is reestablished. c. When temperature is increased for a reaction at equilibrium, the value of K for the reaction will increase. d. When the volume of a reaction container is increased for a system at equilibrium at a given temperature, the reaction will shift left to reestablish equilibrium. e. Addition of a catalyst (a substance that increases the speed of the reaction) has no effect on the equilibrium position.

At \(1100 \mathrm{K}, K_{\mathrm{p}}=0.25\) for the reaction $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ Calculate the equilibrium partial pressures of $\mathrm{SO}_{2}, \mathrm{O}_{2},\( and \)\mathrm{SO}_{3}$ produced from an initial mixture in which \(P_{\mathrm{SO}_{2}}=P_{\mathrm{O}_{2}}=\) 0.50 \(\mathrm{atm}\) and \(P_{\mathrm{so}_{3}}=0 .\) (Hint: If you don't have a graphing calculator, then use the method of successive approximations to solve, as discussed in Appendix \(1.4 . )\)

A sample of gaseous nitrosyl bromide (NOBr) was placed in a container fitted with a frictionless, massless piston, where it decomposed at $25^{\circ} \mathrm{C}$ according to the following equation: $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ The initial density of the system was recorded as 4.495 $\mathrm{g} / \mathrm{L}$ . After equilibrium was reached, the density was noted to be 4.086 \(\mathrm{g} / \mathrm{L}\) . a. Determine the value of the equilibrium constant \(K\) for the reaction. b. If \(\operatorname{Ar}(g)\) is added to the system at equilibrium at constant temperature, what will happen to the equilibrium position? What happens to the value of \(K ?\) Explain each answer.

For the reaction $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \leftrightharpoons 2 \mathrm{CO}(g)$$ \(K_{\mathrm{p}}=2.00\) at some temperature. If this reaction at equilibrium has a total pressure of 6.00 \(\mathrm{atm}\) , determine the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) in the reaction container.

The value of the equilibrium constant \(K\) depends on which of the following (more than one answer may be correct)? a. the initial concentrations of the reactants b. the initial concentrations of the products c. the temperature of the system d. the nature of the reactants and products Explain.
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