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Consider the following exothermic equilibrium (Boudouard reaction) $$2 \mathrm{CO}(g) \rightleftharpoons \mathrm{C}(s)+\mathrm{CO}_{2}(g)$$ How will each of the following changes affect an equilibrium mixture of the three gases: (a) a catalyst is added to the mixture; \((\mathbf{b}) \mathrm{CO}_{2}(g)\) is added to the system; \((\mathbf{c}) \mathrm{CO}(g)\) is added from the system; \((\mathbf{d})\) the reaction mixture is heated; (e) the volume of the reaction vessel is doubled; \((\mathbf{f})\) the total pressure of the system is increased by adding a noble gas?

Short Answer

Expert verified
(a) The equilibrium mixture remains unchanged. (b) The equilibrium shifts to the left, increasing the concentration of CO(g) and decreasing the concentration of C(s) and CO2(g). (c) The equilibrium shifts to the right, increasing the concentration of C(s) and CO2(g) and decreasing the concentration of CO(g). (d) The equilibrium shifts to the left, increasing the concentration of CO(g) and decreasing the concentration of C(s) and CO2(g). (e) The equilibrium shifts to the left, increasing the concentration of CO(g) and decreasing the concentration of C(s) and CO2(g). (f) Adding a noble gas has no effect on the equilibrium position or concentrations of the reactants or products.

Step by step solution

01

a) Addition of a catalyst

Adding a catalyst to the system speeds up both the forward and reverse reactions. However, it does not affect the position of the equilibrium. So in this case, the equilibrium mixture remains unchanged.
02

b) Adding CO2(g)

When CO2 is added to the system, according to Le Chatelier's principle, the equilibrium will shift in the direction that consumes the added substance. In this case, the equilibrium will shift to the left, increasing the concentration of CO(g) and decreasing the concentration of C(s) and CO2(g) until a new equilibrium is established.
03

c) Adding CO(g)

When CO(g) is added to the system, according to Le Chatelier's principle, the equilibrium will shift in the direction that consumes the added substance. In this case, the equilibrium will shift to the right, increasing the concentration of C(s) and CO2(g) and decreasing the concentration of CO(g) until a new equilibrium is established.
04

d) Heating the reaction mixture

As the given reaction is exothermic, heating the mixture will shift the equilibrium towards the endothermic direction (the reverse reaction) to counteract the change. In this case, the equilibrium will shift to the left, increasing the concentration of CO(g) and decreasing the concentration of C(s) and CO2(g) until a new equilibrium is established.
05

e) Doubling the volume of the reaction vessel

Doubling the volume of the reaction vessel reduces the pressure of the system. According to Le Chatelier's principle, the equilibrium will shift in the direction that increases the number of moles of gas. In this case, the equilibrium will shift to the left (as there are 2 moles of CO(g) on the left side and 1 mole of CO2(g) on the right side), increasing the concentration of CO(g) and decreasing the concentration of C(s) and CO2(g) until a new equilibrium is established.
06

f) Increasing the total pressure by adding a noble gas

Adding a noble gas to the system does not change the partial pressures of the original reactants or products, as noble gases do not interact with other chemicals in the reaction. Therefore, adding a noble gas will have no effect on the equilibrium position or the concentrations of the reactants or products.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemistry that helps explain how a system at equilibrium responds to external changes. When an equilibrium system is disturbed, this principle predicts that the system will adjust to counteract that disturbance and reach a new state of balance. This can involve changes such as adjustments in concentration, temperature, or pressure.

For instance, if a reactant or product is added to the system, the equilibrium will shift in the direction that uses up the added component. Conversely, removing a component will shift the equilibrium towards the side that replenishes the removed substance. This behavior is crucial in understanding chemical reactions and predicting their shifts when conditions change.
Exothermic Reactions
In exothermic reactions, energy is released as heat. This is indicated by a negative enthalpy change (\(\Delta H < 0\)). These reactions typically warm their surroundings as they progress.

In the context of equilibrium, if an exothermic reaction is heated, it will shift towards the endothermic side (if possible). This happens because the system attempts to absorb the excess heat, which is a product in the forward reaction of an exothermic process. So, increasing temperature in such systems results in a shift towards the reverse reaction, decreasing the formation of products produced through the forward exothermic reaction.
Reaction Rates
Reaction rates refer to how quickly a chemical reaction occurs. This is influenced by factors like concentration, temperature, pressure, and the presence of a catalyst.

It's essential to understand that while catalysts can speed up both forward and reverse reaction rates, they do not change the equilibrium position. This means that while catalysts are crucial for speeding up the time it takes to reach equilibrium, they do not affect the concentrations of the reactants and products at equilibrium.
  • A higher concentration of reactants typically increases reaction rate.
  • Increased temperature usually speeds up reactions by providing energy to overcome activation barriers.
  • Pressure changes mainly affect reactions involving gases, where increased pressure can favor the direction that shows a decrease in gaseous moles.
Equilibrium Shift
Equilibrium shift describes the movement of an equilibrium position once a change is implemented in the system. Using Le Chatelier's Principle as guidance, the direction of the shift can be predicted by identifying what the system will do to oppose the imposed change.

For example, if additional products are introduced to a system, the equilibrium shifts towards the reactants to consume added products and re-establish balance. Conversely, introducing more reactants causes a shift towards the products.

In gas-phase reactions, changes in pressure play a significant role. If pressure increases (without adding more reactants or products), the equilibrium will tend to shift towards the side with fewer moles of gas to decrease pressure within the system.

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Most popular questions from this chapter

At \(800 \mathrm{~K},\) the equilibrium constant for the reaction \(\mathrm{A}_{2}(g) \rightleftharpoons 2 \mathrm{~A}(g)\) is \(K_{c}=3.1 \times 10^{-4}\). (a) Assuming both forward and reverse reactions are elementary reactions, which rate constant do you expect to be larger, \(k_{f}\) or \(k_{r} ?\) (b) If the value of \(k_{f}=0.27 \mathrm{~s}^{-1}\), what is the value of \(k_{r}\) at \(800 \mathrm{~K} ?\) (c) Based on the nature of the reaction, do you expect the forward reaction to be endothermic or exothermic? (d) If the temperature is raised to \(1000 \mathrm{~K}\), will the reverse rate constant \(k_{r}\) increase or decrease? Will the change in \(k_{r}\) be larger or smaller than the change in \(k_{f}\) ?

At \(900^{\circ} \mathrm{C}, K_{p}=51.2\) for the equilibrium $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ If the pressure of \(\mathrm{NO}(g)\) is half the pressure of \(\mathrm{NOBr}(g)\), what is the equilibrium pressure of \(\mathrm{Br}_{2}(g)\) ?

Bromine and hydrogen react in the gas phase to form hydrogen bromide: \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) .\) The reaction enthalpy is \(\Delta H^{\circ}=-6 \mathrm{~kJ} .\) (a) To increase the equilibrium yield of hydrogen bromide would you use high or low temperature? (b) Could you increase the equilibrium yield of hydrogen bromide by controlling the pressure of this reaction? If so, would high or low pressure favor formation of \(\mathrm{HBr}(g) ?\)

Ozone, \(\mathrm{O}_{3},\) decomposes to molecular oxygen in the stratosphere according to the reaction \(2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g)\). Would an increase in pressure favor the formation of ozone or of oxygen?

Nitric oxide (NO) reacts readily with chlorine gas as follows: $$2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)$$ At \(700 \mathrm{~K},\) the equilibrium constant \(K_{p}\) for this reaction is \(2.6 \times 10^{-3}\). Predict the behavior of each of the following mixtures at this temperature and indicate whether or not the mixtures are at equilibrium. If not, state whether the mixture will need to produce more products or reactants to reach equilibrium. (a) \(P_{\mathrm{NO}}=20.3 \mathrm{kPa}, P_{\mathrm{Cl}_{2}}=20.3 \mathrm{kPa}, R_{\mathrm{NOCl}}=20.3 \mathrm{kPa}\) (b) \(P_{\mathrm{NO}}=25.33 \mathrm{kPa}, P_{\mathrm{Cl}_{2}}=15.2 \mathrm{kPa}, R_{\mathrm{NOCl}}=2.03 \mathrm{kPa}\) (c) \(P_{\mathrm{NO}}=15.2 \mathrm{kPa}, P_{\mathrm{Cl}_{2}}=42.6 \mathrm{kPa}, P_{\mathrm{NOCl}}=5.07 \mathrm{kPa}\)

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