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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
The equilibrium constant for a gas-phase exothermic reaction remains unchanged upon the removal of a reactant (a) or product (b), as well as upon a decrease in volume (c) and the addition of a catalyst (e), due to its dependence solely on temperature. However, when the temperature decreases (d), the equilibrium constant will also decrease since the equilibrium shifts to the right, favoring the exothermic forward reaction.

Step by step solution

01

(a) Removal of a reactant

Removing a reactant from the system will decrease its concentration. According to Le Chatelier's principle, when the concentration of a reactant decreases, the equilibrium will shift to the left to compensate for the change and form more reactants. However, this shift will not affect the value of the equilibrium constant, as it only depends on the temperature and not on the concentrations of the reactants or products.
02

(b) Removal of a product

Removing a product from the system will decrease its concentration. According to the Le Chatelier's principle, when the concentration of a product decreases, the equilibrium will shift to the right to compensate for the change and form more products. Again, this shift will not affect the value of the equilibrium constant, as it only depends on temperature and not on the concentrations of the reactants or products.
03

(c) Decrease in volume

A decrease in volume will result in an increase in pressure, assuming the temperature and the number of moles of gases remain constant. According to Le Chatelier's principle, when the pressure increases, the equilibrium will shift in the direction with fewer gaseous moles to minimize the change. However, this shift in equilibrium will not affect the equilibrium constant because, as mentioned earlier, the equilibrium constant depends only on temperature.
04

(d) Decrease in temperature

Since the reaction is exothermic, a decrease in temperature will favor the forward reaction because it releases heat. According to Le Chatelier's principle, the equilibrium will shift to the right, and more products will be formed to compensate for the decrease in temperature. This change in temperature will affect the value of the equilibrium constant, as it depends on temperature. In this case, the value of the equilibrium constant will decrease as the temperature decreases.
05

(e) Addition of a catalyst

The addition of a catalyst will increase the rate of both the forward and reverse reactions, resulting in a faster approach to equilibrium. However, a catalyst will not affect the value of the equilibrium constant because, as mentioned earlier, the equilibrium constant depends only on temperature and not on the rate at which equilibrium is reached.

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Most popular questions from this chapter

When the following reactions come to equilibrium, does the equilibrium mixture contain mostly reactants or mostly products? (a) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) ; K_{c}=1.5 \times 10^{-10}\) (b) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) ; K_{p}=2.5 \times 10^{9}\)

(a) If \(Q_{c}>K_{c}\), how must the reaction proceed to reach equilibrium? (b) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of \(Q_{c}\) at this point in the reaction?

At \(1200 \mathrm{~K}\), the approximate temperature of automobile exhaust gases (Figure 15.15), \(K_{p}\) for the reaction $$ 2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) $$ is about \(1 \times 10^{-13}\). Assuming that the exhaust gas (total pressure \(1 \mathrm{~atm}\) ) contains \(0.2 \% \mathrm{CO}, 12 \% \mathrm{CO}_{2}\), and \(3 \% \mathrm{O}_{2}\) by volume, is the system at equilibrium with respect to the \(\mathrm{CO}_{2}\) reaction? Based on your conclusion, would the \(\mathrm{CO}\) concentration in the exhaust be decreased or increased by a catalyst that speeds up the \(\mathrm{CO}_{2}\) reaction? Recall that at a fixed pressure and temperature, volume \(\%=\mathrm{mol} \%\).

Which of the following reactions lies to the right, favoring the formation of products, and which lies to the left, favoring formation of reactants? (a) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) ; K_{p}=5.0 \times 10^{12}\) (b) \(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) ; K_{c}=5.8 \times 10^{-18}\)

Consider the following equilibrium: $$ 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g) \quad K_{c}=1.08 \times 10^{7} \text { at } 700{ }^{\circ} \mathrm{C} $$ (a) Calculate \(K_{p-}\) (b) Does the equilibrium mixture contain mostly \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) or mostly \(\mathrm{H}_{2} \mathrm{~S}\) ? (c) Calculate the value of \(K_{c}\) if you rewrote the equation \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{~S}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)\).
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