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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

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

Step by step solution

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 \(K_{\mathrm{c}}\).

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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

Consider the equilibrium: $$ 2 \mathrm{I}(g) \rightleftarrows \mathrm{I}_{2}(g) $$ What would be the effect on the position of equilibrium of (a) increasing the total pressure on the system by decreasing its volume, (b) adding \(\mathrm{I}_{2}\) to the reaction mixture, and (c) decreasing the temperature?

At equilibrium, the pressure of the reacting mixture $$\mathrm{CaCO}_{3}(s) \rightleftarrows \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ is 0.105 atm at \(350^{\circ} \mathrm{C}\). Calculate \(K_{P}\) and \(K_{c}\) for this reaction.

At \(20^{\circ} \mathrm{C},\) the vapor pressure of water is \(0.0231 \mathrm{~atm} .\) Calculate \(K_{P}\) and \(K_{\mathrm{c}}\) for the process: $$\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{H}_{2} \mathrm{O}(g)$$

For which of the following reactions is \(K_{\mathrm{c}}\) equal to \(K_{P}\) ? For which can we not write a \(K_{P}\) expression? (a) \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftarrows 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) (b) \(\mathrm{CaCO}_{3}(s) \rightleftarrows \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\) (c) \(\mathrm{Zn}(s)+2 \mathrm{H}^{+}(a q) \rightleftarrows \mathrm{Zn}^{2+}(a q)+\mathrm{H}_{2}(g)\) (d) \(\mathrm{PCl}_{3}(g)+3 \mathrm{NH}_{3}(g) \rightleftarrows 3 \mathrm{HCl}(g)+\mathrm{P}\left(\mathrm{NH}_{2}\right)_{3}(g)\) (e) \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \rightleftarrows \mathrm{NH}_{4} \mathrm{Cl}(s)\) (f) \(\mathrm{NaHCO}_{3}(s)+\mathrm{H}^{+}(a q) \rightleftarrows \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)+\) \(\mathrm{Na}^{+}(a q)\) (g) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftarrows 2 \mathrm{HF}(g)\) (h) \(\mathrm{C}(\) graphite \()+\mathrm{CO}_{2}(g) \rightleftarrows 2 \mathrm{CO}(g)\)

Consider the equilibrium system \(3 \mathrm{~A} \rightleftarrows \mathrm{B}\). Sketch the changes in the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) over time for the following situations: (a) initially only \(\mathrm{A}\) is present, (b) initially only B is present, (c) initially both A and \(B\) are present (with A in higher concentration). In each case, assume that the concentration of \(\mathrm{B}\) is higher than that of \(\mathrm{A}\) at equilibrium.

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