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Consider the following equilibrium systems: (a) \(\mathrm{A} \rightleftarrows 2 \mathrm{~B} \quad \Delta H^{\circ}=20.0 \mathrm{~kJ} / \mathrm{mol}\) (b) \(\mathrm{A}+\mathrm{B} \rightleftarrows \mathrm{C} \quad \Delta H^{\circ}=-5.4 \mathrm{~kJ} / \mathrm{mol}\) (c) \(\mathrm{A} \rightleftarrows \mathrm{B} \quad \Delta H^{\circ}=0.0 \mathrm{~kJ} / \mathrm{mol}\) Predict the change in the equilibrium constant \(K_{\mathrm{c}}\) that would occur in each case if the temperature of the reacting system were raised.

Short Answer

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(a) Increase; (b) Decrease; (c) No change.

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01

Understand the Effect of Temperature on Equilibrium

The effect of temperature on the equilibrium constant is dictated by Le Chatelier's Principle and the sign of the enthalpy change (\(\Delta H^{\circ}\)) of the reaction. A positive \(\Delta H^{\circ}\) means the reaction is endothermic, absorbing heat, and a negative \(\Delta H^{\circ}\) means the reaction is exothermic, releasing heat. Raising the temperature will favor the endothermic direction and oppose the exothermic direction.
02

Analyze Reaction (a)

For reaction (a): \(\mathrm{A} \rightleftarrows 2 \mathrm{~B} \) with \(\Delta H^{\circ}=20.0\, \mathrm{~kJ/mol}\), the reaction is endothermic. Increasing the temperature will favor the production of \(\mathrm{B}\), increasing the equilibrium constant \(K_{\mathrm{c}}\).
03

Analyze Reaction (b)

For reaction (b): \(\mathrm{A} + \mathrm{B} \rightleftarrows \mathrm{C} \) with \(\Delta H^{\circ}=-5.4\, \mathrm{~kJ/mol}\), the reaction is exothermic. Increasing the temperature will favor the reverse reaction, decreasing the equilibrium constant \(K_{\mathrm{c}}\).
04

Analyze Reaction (c)

For reaction (c): \(\mathrm{A} \rightleftarrows \mathrm{B} \) with \(\Delta H^{\circ}=0.0\, \mathrm{~kJ/mol}\), the reaction does not absorb or release heat. Therefore, changing the temperature has no effect on the equilibrium constant \(K_{\mathrm{c}}\).

Key Concepts

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

Equilibrium Constant
The equilibrium constant, denoted as \(K_c\), is a key factor in understanding chemical reactions at equilibrium. It provides a measure of the ratio of the concentrations of products to reactants at equilibrium. This ratio is maintained at a constant value for a given reaction at a fixed temperature.
When you raise or lower the temperature, the equilibrium constant can change, depending on whether the reaction is endothermic or exothermic. Therefore, the value of \(K_c\) is temperature-dependent, meaning that as temperature changes, \(K_c\) responds accordingly to establish a new equilibrium position. Understanding \(K_c\) will help you predict how changes in conditions will affect a chemical reaction at equilibrium.
Endothermic Reaction
An endothermic reaction is a chemical reaction that absorbs heat from its surroundings. These reactions have a positive enthalpy change \((\Delta H^\circ > 0)\). For endothermic reactions, an increase in temperature tends to favor the forward reaction, as adding heat provides the necessary energy to proceed.
This means that raising temperature will increase the equilibrium constant for endothermic reactions. In essence, as more heat means providing more energy, the system naturally shifts towards producing more products, thus increasing \(K_c\). It's like giving the reaction a little push to keep moving in the direction of product formation.
Exothermic Reaction
In contrast to endothermic reactions, exothermic reactions release heat into their surroundings. These have a negative enthalpy change \((\Delta H^\circ < 0)\). When you increase the temperature in an exothermic reaction, the system works to counteract this change by favoring the reverse reaction.
This results in a decrease in the equilibrium constant. Essentially, the system is trying to "cool down" by reducing the formation of products, thereby lowering \(K_c\). Remember, exothermic reactions dislike added heat, and so, their natural response is to shift in such a way as to use up the extra heat.
Enthalpy Change
Enthalpy change, \(\Delta H^\circ\), tells us about the heat absorbed or released during a chemical reaction. In discussing chemical equilibrium, enthalpy change provides insights into the reaction nature. It acts as a critical marker to predict how equilibrium will respond to temperature changes.
  • In an endothermic reaction \((\Delta H^\circ > 0)\), adding heat shifts the reaction towards the products. Hence, increasing \(K_c\).
  • In an exothermic reaction \((\Delta H^\circ < 0)\), adding heat pushes the reaction towards the reactants. Therefore, decreasing \(K_c\).
Understanding enthalpy change enables you to predict how a system at equilibrium reacts to external temperature variations, giving you the means to control chemical reactions more effectively.

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Most popular questions from this chapter

A quantity of 0.20 mole of carbon dioxide was heated to a certain temperature with an excess of graphite in a closed container until the following equilibrium was reached: $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftarrows 2 \mathrm{CO}(g) $$ Under these conditions, the average molar mass of the gases was \(35 \mathrm{~g} / \mathrm{mol}\). (a) Calculate the mole fractions of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\). (b) What is \(K_{P}\) if the total pressure is 11 atm? (Hint: The average molar mass is the sum of the products of the mole fraction of each gas and its molar mass.)

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.

A 2.50 -mol sample of \(\mathrm{NOCl}\) was initially in a \(1.50-\mathrm{L}\) reaction chamber at \(400^{\circ} \mathrm{C}\). After equilibrium was established, it was found that 28.0 percent of the \(\mathrm{NOCl}\) had dissociated: $$ 2 \mathrm{NOCl}(g) \rightleftarrows 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.

Consider the following equilibrium reaction in a closed container: $$ \mathrm{CaCO}_{3}(s) \rightleftarrows \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ What will happen if (a) the volume is increased, (b) some \(\mathrm{CaO}\) is added to the mixture, \((\mathrm{c})\) some \(\mathrm{CaCO}_{3}\) is removed, \(\left(\right.\) d) some \(\mathrm{CO}_{2}\) is added to the mixture, (e) a few drops of an \(\mathrm{NaOH}\) solution are added to the mixture, (f) a few drops of an HCl solution are added to the mixture (ignore the reaction between \(\mathrm{CO}_{2}\) and water), (g) temperature is increased?

Industrially, sodium metal is obtained by electrolyzing molten sodium chloride. The reaction at the cathode is \(\mathrm{Na}^{+}+e^{-} \longrightarrow \mathrm{Na}\). We might expect that potassium metal would also be prepared by electrolyzing molten potassium chloride. However, potassium metal is soluble in molten potassium chloride and therefore is hard to recover. Furthermore, potassium vaporizes readily at the operating temperature, creating hazardous conditions. Instead, potassium is prepared by the distillation of molten potassium chloride in the presence of sodium vapor at \(892^{\circ} \mathrm{C}:\) $$\mathrm{Na}(g)+\mathrm{KCl}(l) \rightleftarrows \mathrm{NaCl}(l)+\mathrm{K}(g)$$ In view of the fact that potassium is a stronger reducing agent than sodium, explain why this approach works. (The boiling points of sodium and potassium are \(892^{\circ} \mathrm{C}\) and \(770^{\circ} \mathrm{C}\), respectively.)

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