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Chapter 15: Problem 63

How do the following changes affect the value of the equilibrium constant for a gas-phase exothermic reaction: (a) removal of a reactant, (b) removal of a product, (c) decrease in the volume, (d) decrease in the temperature, (e) addition of a catalyst?

Short Answer

Expert verified
In summary, for a gas-phase exothermic reaction: (a) removal of a reactant and (b) removal of a product affect the position of the equilibrium but not the value of the equilibrium constant K; (c) a decrease in volume affects the position of the equilibrium but not the value of K; (d) a decrease in temperature causes the equilibrium to shift towards the reactants and affects the value of K, decreasing it; (e) the addition of a catalyst helps the reaction reach equilibrium faster without affecting the position of the equilibrium or the value of K.

Step by step solution

01

a) Removal of a reactant

When a reactant is removed, the concentration of that particular reactant decreases. According to Le Chatelier's Principle, the equilibrium will shift in the direction that opposes the change, in this case, shifting towards the reactants. However, this shift does not affect the value of the equilibrium constant (K), as it remains constant at a given temperature. Removing a reactant only affects the position of the equilibrium, not the equilibrium constant.
02

b) Removal of a product

When a product is removed from the system, its concentration decreases. According to Le Chatelier's Principle, the equilibrium will shift in the direction that opposes the change, meaning it will shift towards the products. Again, this shift does not affect the value of the equilibrium constant (K), which remains constant at a given temperature. Removing a product only affects the position of the equilibrium, not the equilibrium constant.
03

c) Decrease in volume

When the volume decreases, the pressure inside the container increases. According to Le Chatelier's Principle, the equilibrium will shift in the direction that would decrease the pressure. In a gaseous reaction, the equilibrium will shift towards the side with fewer moles of gas. However, the equilibrium constant (K) remains unaffected by this change. It is important to note that a change in volume only affects the position of the equilibrium, not the equilibrium constant.
04

d) Decrease in temperature

For an exothermic reaction, a decrease in temperature will cause the equilibrium to shift in the direction that produces heat, meaning it will shift towards the reactants. This is because the reaction releases heat, making the reverse reaction endothermic, which absorbs the heat produced by the decrease in temperature. In this case, the equilibrium constant (K) will also change since it is affected by the change in temperature. For an exothermic reaction, when the temperature decreases, the value of K decreases as well.
05

e) Addition of a catalyst

A catalyst is a substance that accelerates the rate of a chemical reaction without undergoing any permanent chemical change. Catalysts help to speed up both the forward and reverse reactions by providing an alternative reaction pathway with a lower activation energy for the reaction. However, catalysts do not affect the position of the equilibrium or the value of the equilibrium constant (K). The addition of a catalyst will only help the reaction reach equilibrium faster, without changing the equilibrium concentrations of the reactants and products.

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Most popular questions from this chapter

Consider the reaction \(10_{4}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}(a q) ; K_{c}=3.5 \times 10^{-2} .\) If you start with 25.0 \(\mathrm{mL}\) of a 0.905 \(\mathrm{M}\) solution of \(\mathrm{NaIO}_{4},\) and then dilute it with water to 500.0 \(\mathrm{mL}\) , what is the concentration of \(\mathrm{H}_{4} \mathrm{IO}_{6}\) at equilibrium?

Mercury(l) oxide decomposes into elemental mercury and elemental oxygen: \(2 \mathrm{Hg}_{2} \mathrm{O}(s) \rightleftharpoons 4 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) (a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) Suppose you run this reaction in a solvent that dissolves elemental mercury and elemental oxygen. Rewrite the equilibrium- constant expression in terms of molarities for the reaction, using (solv) to indicate solvation.

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 \(\mathrm{Br}_{2}(g) ?\)

Consider the equilibrium $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ Calculate the equilibrium constant \(K_{p}\) for this reaction, given the following information (at 298 \(\mathrm{K} )\) : \begin{equation} \begin{array}{l}{2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \operatorname{NOBr}(g) \quad K_{c}=2.0} \\ {2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \quad K_{c}=2.1 \times 10^{30}}\end{array} \end{equation}

\(\mathrm{At} 900 \mathrm{K},\) the following reaction has \(K_{p}=0.345 :\) $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ In an equilibrium mixture the partial pressures of \(S \mathrm{O}_{2}\) and \(\mathrm{O}_{2}\) are 0.135 atm and 0.455 atm, respectively. What is the equilibrium partial pressure of \(\mathrm{SO}_{3}\) in the mixture?
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