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What is the effect on the equilibrium position if an exothermic reaction is carried out at a higher temperature? Does the net amount of product increase or decrease? Does the value of the equilibrium constant change if the temperature is increased? If so, how does it change?

Short Answer

Expert verified
For an exothermic reaction, when the temperature is increased, the equilibrium position shifts towards the reactants' side, decreasing the net amount of product. Moreover, the value of the equilibrium constant also decreases with increasing temperature, as the reaction is favored at lower temperatures.

Step by step solution

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1. Introduction to chemical equilibrium and exothermic reactions

In a chemical reaction, the equilibrium is reached when the rates of forward and reverse reactions are equal. In an exothermic reaction, heat is released as a product. According to Le Chatelier's Principle, any change in the reaction conditions (such as temperature, pressure, or concentration) will cause a shift in the equilibrium position to counteract the change.
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2. Effect of temperature increase on the equilibrium position of an exothermic reaction

When the temperature of an exothermic reaction is increased, the system will try to absorb the excess heat by shifting the equilibrium position towards the side that consumes heat. Since an exothermic reaction releases heat, the equilibrium will shift towards the reactants' side to reduce the net amount of heat produced.
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3. Change in the net amount of product

As the equilibrium position shifts towards the reactants' side due to the increased temperature, the concentration of products will decrease. Consequently, the net amount of product will also decrease.
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4. Change in the value of the equilibrium constant

The equilibrium constant, K, is related to the temperature and reaction's Gibbs free energy. When increasing the temperature, the value of the equilibrium constant will also change. For an exothermic reaction, increasing the temperature will decrease the equilibrium constant. This is because the reaction is favored at lower temperatures where more products are formed, and it has an easier time releasing the excess heat.
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5. Conclusion

In conclusion, when an exothermic reaction is carried out at a higher temperature, the equilibrium position shifts towards the reactants' side, and the net amount of product decreases. The value of the equilibrium constant also decreases due to the increased temperature.

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Key Concepts

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

Exothermic Reactions
In a chemical reaction, energy changes play a vital role, especially in terms of heat exchange. Exothermic reactions are processes that release heat, meaning that the products of the reaction have less energy than the reactants. This release of energy is what makes exothermic reactions fascinating and commonly observed in everyday phenomena, such as combustion, which powers engines, and even your body generating heat from food metabolism.
When considering exothermic reactions, it's essential to recognize that heat is effectively a product of the reaction. This characteristic influences how the reaction responds to changes in conditions, particularly temperature. As part of understanding this, heat is counted alongside other products in a chemical equation when assessing equilibrium conditions.
Le Chatelier's Principle
Le Chatelier's Principle is a helpful guideline in predicting how a chemical equilibrium will shift when the conditions of a reaction are altered. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to neutralize the effect of the disturbance. Le Chatelier's Principle can be applied to various parameters that affect chemical balance:
  • Temperature
  • Pressure
  • Concentration
When temperature increases in an exothermic reaction, the system will try to counteract the change by absorbing the extra heat. It achieves this by shifting the equilibrium towards the reactants, where heat will be absorbed, thereby reducing excess heat generated. This thoughtful adjustment helps maintain a balance within the chemical system, aligning with nature's preference for equilibrium.
Equilibrium Constant
The equilibrium constant, often denoted by the symbol \( K \), quantifies the balance point of a reaction at a given temperature. It is derived from the concentration of products and reactants at equilibrium and gives insight into the favored direction of the reaction at that temperature. For exothermic reactions, the equilibrium constant decreases as the temperature rises. This is because higher temperatures favor the reverse of an exothermic reaction, where reactants are formed rather than products.When temperature increases for an exothermic reaction, the equilibrium shifts to favor the formation of reactants, reducing the concentration of products. This results in a lower value of \( K \). Understanding \( K \) is crucial for determining how temperature changes affect the yield of a reaction, further showcasing the dynamic yet predictable nature of chemical processes.

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Most popular questions from this chapter

The solubility product constant, \(K_{\mathrm{sp}},\) for calcium carbonate at room temperature is approximately \(3.0 \times 10^{-9}\). Calculate the solubility of \(\mathrm{CaCO}_{3}\) in grams per liter under these conditions.

\(K_{\text {sp }}\) for copper(II) hydroxide, \(\mathrm{Cu}(\mathrm{OH})_{2},\) has a value \(2.2 \times 10^{-20}\) at 25 " \(\mathrm{C}\). Calculate the solubility of copper(II) hydroxide in \(\mathrm{mol} / \mathrm{L}\) and \(\mathrm{g} / \mathrm{L}\) at \(25 \mathrm{C}\)

At high temperatures, elemental nitrogen and oxygen react with each other to form nitrogen monoxide. $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ Suppose the system is analyzed at a particular temperature, and the equilibrium concentrations are found to be \(\left[\mathrm{N}_{2}\right]=0.041 \mathrm{M},\left[\mathrm{O}_{2}\right]=0.0078 M,\) and \([\mathrm{NO}]=4.7 \times 10^{-4} M .\) Calculate the value of \(K\) for the reaction.

Suppose the reaction system $$ \mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ has already reached equilibrium. Predict the effect of each of the following changes on the position of the equilibrium. Tell whether the equilibrium will shift to the right, will shift to the left, or will not be affected. a. Any liquid water present is removed from the system. b. \(\mathrm{CO}_{2}\) is added to the system by dropping a chunk of dry ice into the reaction vessel. c. The reaction is performed in a metal cylinder fitted with a piston, and the piston is compressed to decrease the total volume of the system. d. Additional \(\mathrm{O}_{2}(g)\) is added to the system from a cylinder of pure \(\mathrm{O}_{2}\).

Write the equilibrium expression for each of the following reactions. a. \(\mathrm{N}_{2}(g)+3 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NCl}_{3}(g)\) b. \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\) c. \(\mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{H}_{4}(g)\)

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