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Does the addition of a catalyst have any effects on the position of an equilibrium?

Short Answer

Expert verified
A catalyst does not affect the equilibrium position.

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01

Understand the Function of a Catalyst

A catalyst speeds up the rate of a chemical reaction by providing an alternative pathway with a lower activation energy. This means that the reactions can proceed faster in both directions (forward and backward) in a reversible reaction.
02

Define Chemical Equilibrium

Chemical equilibrium is established in a reversible reaction when the rate of the forward reaction equals the rate of the backward reaction, leading to constant concentrations of the reactants and products over time.
03

Evaluate Catalyst's Role in Equilibrium

While a catalyst accelerates the speed at which equilibrium is reached, it does not change the concentrations of reactants and products at equilibrium. Therefore, the catalyst does not shift the position of the equilibrium.

Key Concepts

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

Catalysis
A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process. It works by providing an alternative reaction pathway with a lower activation energy. By lowering activation energy, the catalyst allows the reaction to proceed more easily, thus speeding up both the forward and reverse reactions in reversible reactions. This is because the catalyst affects the reaction kinetics, helping molecules to collide more effectively and transform into products more rapidly on the molecular level.
Since the catalyst itself is not consumed, it can be used repeatedly, making this process efficient. Importantly, while a catalyst can speed up the rate at which equilibrium is achieved, it does not influence the equilibrium position or change the concentrations of reactants and products. Thus, the catalyst only makes the reaction faster, not altering the balance between reactants and products.
Chemical Equilibrium
Chemical equilibrium occurs in a reversible reaction when the rate of the forward reaction equals the rate of the backward reaction. This leads to a state where the concentrations of reactants and products remain constant over time. Equilibrium is dynamic; even though the concentrations remain constant, the reactions themselves are still occurring at the molecular level.
At equilibrium, it might seem like the reaction has stopped, but it's actually a balance between the two opposing processes. This state can be represented by the equilibrium constant, which is determined by the ratio of the concentrations of products to reactants. However, the equilibrium position is not altered by the presence of a catalyst; it remains dependent on factors like temperature and the initial concentrations of reactants and products.
Activation Energy
Activation energy is the minimum amount of energy required for a chemical reaction to occur. It is also the energy barrier that reactants need to overcome to be converted into products. The concept of activation energy is crucial in understanding how catalysts work.
A catalyst lowers the activation energy needed for the reaction, meaning that more reactant molecules will have enough energy to react when a catalyst is present. This increases the reaction rate, as more molecules can surpass the energy barrier at a given time. By providing an alternative reaction pathway with lower activation energy, catalysts make reactions proceed faster without being part of the final products, contributing significantly to efficient chemical processes.
Reaction Rates
Reaction rates describe how quickly or slowly a chemical reaction occurs. They are influenced by several factors, including the presence of a catalyst, temperature, concentration of reactants, and surface area of reactants.
  • With a catalyst, reaction rates increase because the catalyst lowers the activation energy needed for the reaction to proceed.
  • Temperature can also affect reaction rates, as it provides energy to the reacting molecules, increasing their movement and the likelihood of collisions.
  • Higher concentrations of reactants typically lead to higher reaction rates since more molecules are available to collide and react.
  • The surface area of reactants matters as well, especially in heterogeneous reactions where reactants are in different phases.
While these factors affect how quickly equilibrium is reached, they do not alter the equilibrium position, which is determined by the energy dynamics between reactants and products.

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Most popular questions from this chapter

The equilibrium constant \(\left(K_{P}\right)\) for the formation of the air pollutant nitric oxide (NO) in an automobile engine $$\begin{array}{l} \text { at } 530^{\circ} \mathrm{C} \text { is } 2.9 \times 10^{-11}: \\ \qquad \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g) \end{array}$$ (a) Calculate the partial pressure of NO under these conditions if the partial pressures of nitrogen and oxygen are 3.0 and 0.012 atm, respectively. (b) Repeat the calculation for atmospheric conditions where the partial pressures of nitrogen and oxygen are 0.78 and 0.21 atm and the temperature is \(25^{\circ} \mathrm{C}\). (The \(K_{P}\) for the reaction is \(4.0 \times 10^{-31}\) at this temperature.) (c) Is the formation of NO endothermic or exothermic? (d) What natural phenomenon promotes the formation of NO? Why?

Photosynthesis can be represented by: $$\begin{aligned} 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) & \rightleftarrows \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \\ \Delta H^{\circ} &=2801 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$$ Explain how the equilibrium would be affected by the following changes: (a) partial pressure of \(\mathrm{CO}_{2}\) is increased, (b) \(\mathrm{O}_{2}\) is removed from the mixture, (c) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (glucose) is removed from the mixture, (d) more water is added, (e) a catalyst is added, (f) temperature is decreased.

The equilibrium constant \(K_{P}\) for the reaction: $$ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$ the same temperature? (b) The very small value of \(K_{P}\) (and \(K_{\mathrm{c}}\) ) indicates that the reaction overwhelmingly favors the formation of water molecules. Explain why, despite this fact, a mixture of hydrogen and oxygen gases can be kept at room temperature without any change.

A mixture containing 3.9 moles of \(\mathrm{NO}\) and 0.88 mole of \(\mathrm{CO}_{2}\) was allowed to react in a flask at a certain temperature according to the equation: $$ \mathrm{NO}(g)+\mathrm{CO}_{2}(g) \rightleftarrows \mathrm{NO}_{2}(g)+\mathrm{CO}(g) $$ At equilibrium, 0.11 mole of \(\mathrm{CO}_{2}\) was present. Calculate the equilibrium constant \(K_{\mathrm{c}}\) of this reaction.

In the uncatalyzed reaction: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g) $$ the pressure of the gases at equilibrium are \(P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.377\) atm and \(P_{\mathrm{NO}_{2}}=1.56\) atm at \(100^{\circ} \mathrm{C}\). What would happen to these pressures if a catalyst were added to the mixture?

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