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Chapter 12: Problem 98

For the following endothermic reaction at equilibrium: $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ which of the following changes will increase the value of \(K ?\) a. increasing the temperature b. decreasing the temperature c. removing \(\mathrm{SO}_{3}(g)\) (constant \(T\) ) d. decreasing the volume (constant \(T\) ) e. adding Ne(g) (constant \(T\) ) f. adding \(\mathrm{SO}_{2}(g)\) (constant \(T\) ) g. adding a catalyst (constant \(T\) )

Short Answer

Expert verified
In conclusion, only option a (increasing the temperature) will increase the value of K for the given endothermic reaction at equilibrium.

Step by step solution

01

Analyze the effect of temperature changes

According to Le Chatelier's Principle, if an endothermic reaction is at equilibrium and the temperature is increased, the reaction will shift towards the products (right) to counteract the increase in temperature. Therefore, the value of K will increase. On the other hand, if the temperature is decreased, the reaction will shift towards the reactants (left), and thus the value of K will decrease. Hence, increasing the temperature (option a) will increase the value of K, while decreasing the temperature (option b) will not.
02

Analyze the effect of concentration changes

According to Le Chatelier's principle, when the concentration of a reactant or product changes, the reaction will shift towards the side that opposes the change. Removing SO3 (option c) will cause the reaction to shift towards the reactants, but this change in the equilibrium position won't affect the value of K. Adding SO2 (option f) will cause the reaction to shift towards the products, but again, this change in the equilibrium position won't affect the value of K. Both options c and f will alter the equilibrium position, but they will not change the value of K.
03

Analyze the effect of volume and pressure changes

Decreasing the volume (option d) at constant temperature will result in an increased pressure. According to Le Chatelier's Principle, the system will shift towards the side with fewer moles of gas to counteract the pressure increase. Since the number of moles of gas on both sides of the equilibrium is the same (i.e., 2 moles on both sides), the system will not shift in either direction, and the value of K will remain constant.
04

Analyze the effect of adding an inert gas and catalyst

Adding an inert gas (option e) at constant temperature and volume won't affect the partial pressures or concentrations of the reactants and products, so the equilibrium position will remain unchanged, and the value of K will not be affected. Adding a catalyst (option g), on the other hand, will increase the rate of both the forward and reverse reactions equally, but it will not affect the relative amounts of reactants and products at equilibrium, and thus the value of K will remain constant. In conclusion, only option a (increasing the temperature) will increase the value of K for the given endothermic reaction at equilibrium.

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Most popular questions from this chapter

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

Peptide decomposition is one of the key processes of digestion, where a peptide bond is broken into an acid group and an amine group. We can describe this reaction as follows: \(\text { Peptide }(a q)+\mathrm{H}_{2} \mathrm{O}(t) \rightleftharpoons \text { acid group }(a q)+\text { amine group }(a q)\)

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 \(\mathrm{Ar}(g)\) 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.

A gaseous material \(\mathrm{XY}(g)\) dissociates to some extent to produce \(\mathrm{X}(g)\) and \(\mathrm{Y}(g):\).$$\mathrm{XY}(g) \rightleftharpoons \mathrm{X}(g)+\mathrm{Y}(g)$$.A 2.00 -g sample of XY (molar mass \(=165 \mathrm{g} / \mathrm{mol}\) ) is placed in a container with a movable piston at \(25^{\circ} \mathrm{C}\). The pressure is held constant at 0.967 atm. As XY begins to dissociate, the piston moves until 35.0 mole percent of the original XY has dissociated and then remains at a constant position. Assuming ideal behavior, calculate the density of the gas in the container after the piston has stopped moving, and determine the value of \(K\) for this reaction of \(25^{\circ} \mathrm{C}\).

Given \(K=3.50\) at \(45^{\circ} \mathrm{C}\) for the reaction $$\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)$$ and \(K=7.10\) at \(45^{\circ} \mathrm{C}\) for the reaction $$2 \mathrm{A}(g)+\mathrm{D}(g) \rightleftharpoons \mathrm{C}(g)$$what is the value of \(K\) at the same temperature for the reaction $$ \mathrm{C}(g)+\mathrm{D}(g) \rightleftharpoons 2 \mathrm{B}(g)$$.What is the value of \(K_{\mathrm{p}}\) at \(45^{\circ} \mathrm{C}\) for the reaction? Starting with 1.50 atm partial pressures of both \(\mathrm{C}\) and \(\mathrm{D},\) what is the mole fraction of B once equilibrium is reached?
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