Use Le Chatelier's principle to predict the effect that the change given in the first column of the following table has on the quantity in the second column for the following equilibrium system: $$ \begin{gathered} 5 \mathrm{CO}(\mathrm{g})+\mathrm{I}_{2} \mathrm{O}_{5}(\mathrm{~s}) \rightleftharpoons \mathrm{I}_{2}(\mathrm{~g})+5 \mathrm{CO}_{2}(\mathrm{~g}) \\\ \Delta H^{\circ}=-1175 \mathrm{~kJ} \end{gathered} $$

Short Answer

Expert verified
For an exothermic reaction with \( \Delta H^\circ = -1175 \text{ kJ} \), increasing temperature shifts the equilibrium to the left, while decreasing temperature shifts it to the right. Changes in pressure have no effect due to equal moles of gas on both sides, and changes in concentration will shift the equilibrium to the side that counteracts the change.

Step by step solution

01

Understand Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. This principle can be used to predict the effects of changes in concentration, pressure, or temperature on the position of equilibrium in a chemical reaction.
02

Identifying the Type of Change

The problem description did not specify which type of change to the system is being made. Typical changes include changes in concentration, pressure, temperature, or the addition of a catalyst. For each change, the position of equilibrium will shift to either the right or left, increasing the production of either the products or the reactants, respectively. Without specific details on the change, a general analysis on the equilibrium adjustment to temperature change is to be provided, since the reaction's enthalpy change, \( \Delta H^\circ \), is given.
03

Predict Effects of Temperature Change

Since the enthalpy change \( \Delta H^\circ \) for the reaction is negative, the reaction is exothermic. Increasing temperature adds heat to the system, which according to Le Chatelier's Principle, will shift the equilibrium to the left, favoring the reactants. Decreasing temperature has the opposite effect; it removes heat causing the equilibrium to shift to the right, favoring the formation of products.
04

Predict Effects of Pressure Change on Gases

If the pressure of the system is increased, the equilibrium will shift towards the side with fewer moles of gas. In this case, there are initially 5 moles of CO(g) and 1 mole of I2O5(s), and the products are I2(g) and 5 CO2(g). Since solids do not affect the pressure, we look at the difference in gas moles only. There is no change in the number of gas moles (5 on both sides), so pressure change will not affect the position of equilibrium in this reaction.
05

Predict Effects of Concentration Change

If the concentration of CO(g) is increased, the equilibrium will shift to the right to increase the production of I2(g) and CO2(g). Conversely, increasing the concentration of either I2(g) or CO2(g) would shift the equilibrium to the left to produce more CO(g) and I2O5(s). If I2O5(s) is added, there will likely be little to no shift in equilibrium because pure solids do not have concentrations that can change.

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Key Concepts

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

Understanding Chemical Equilibrium
Chemical equilibrium is a state in which the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products over time. It is vital to recognize that equilibrium does not mean that the reactants and products are present in equal amounts, but rather that their ratios are constant.

Consider a simple reversible reaction where the reactants A and B form products C and D:
  • Forward reaction: A + B → C + D
  • Reverse reaction: C + D → A + B
At equilibrium, the rate at which A and B react to form C and D is the same as the rate at which C and D revert back to A and B. This balance of rates ensures that the overall concentration of each species remains constant, though they may continue to react on a molecular level.
Exothermic Reactions and Heat
Exothermic reactions are characterized by the release of energy, usually in the form of heat, to the surroundings. The reaction enthalpy, denoted as \( \Delta H^\circ \), is negative for exothermic processes, indicating that energy is exiting the system.

For example, when you feel warmth from a chemical hand warmer, it is the result of an exothermic reaction occurring inside. This release of thermal energy is spontaneous and converts the reactants into products while emitting heat:
  • Reactants → Products + Heat
The generated heat can influence the surrounding environment and, in a closed system, can also affect the reaction itself if temperature changes occur.
Equilibrium Shifts from Temperature Changes
According to Le Chatelier's Principle, if the temperature of an equilibrium system is altered, the system will adjust to minimize the impact of this change. In the context of exothermic reactions, an increase in temperature injects additional heat, which the system counteracts by shifting the equilibrium to favor the reactants, effectively absorbing the excess heat.

Decreasing the temperature, on the other hand, removes heat from the system. As a result, the equilibrium will adjust by shifting toward the products, as the exothermic reaction releases heat, compensating for the lost thermal energy. This shift is predictable: if you cool down an exothermic reaction, expect more products to form as the system tries to maintain its balanced state.
Reaction Enthalpy and Equilibrium
Reaction enthalpy (\( \Delta H^\circ \)) plays a crucial role in determining how a chemical equilibrium responds to changes in temperature. A negative \( \Delta H^\circ \) signifies an exothermic reaction. When such a reaction is subjected to a temperature increase, Le Chatelier's Principle indicates that the system will shift to absorb the added heat, thus moving the equilibrium toward the reactants. Conversely, decreasing the temperature causes the equilibrium to shift toward the products to release heat and raise the system's temperature.

It is important to remember that while \( \Delta H^\circ \) provides information about the heat change during the reaction, it also predicts the direction of the equilibrium shift in response to temperature fluctuations. Recognizing the sign of \( \Delta H^\circ \) is thus essential for anticipating how equilibrium will re-establish itself under changing conditions.

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Most popular questions from this chapter

Explain why the following equilibria are heterogencous and write the reaction quoticnt \(Q\) for each one. (a) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})=\mathrm{NH}_{3}(\mathrm{~g})+\mathrm{HCl}(\mathrm{g})\) (b) \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{OH}_{2} \mathrm{O}\) (s) \(\rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{~s})+10 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(2 \mathrm{KCO}_{3}\) (s) \(\Rightarrow 2 \mathrm{KCl}\) (s) \(+3 \mathrm{O}_{2}\) (g)

Predict whether each of the following equilibria will shift toward products or reactants with a temperature increase. (a) \(\mathrm{N}_{2} \mathrm{O}_{4}(g)=2 \mathrm{NO}_{2}(g), \Delta H^{\circ}=+57 \mathrm{~kJ}\) (b) \(\mathrm{X}_{2}(\mathrm{~g})=2 \mathrm{X}(\mathrm{g})\), where \(\mathrm{X}\) is a halogen (c) \(\mathrm{Ni}(\mathrm{s})+4 \mathrm{CO}(\mathrm{g})=\mathrm{Ni}(\mathrm{CO})_{4}(\mathrm{~g}), \Delta \mathrm{H}^{+}=-161 \mathrm{~kJ}\) (d) \(\mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), \(\Delta H^{\circ}=-90 \mathrm{~kJ}\)

At \(1565 \mathrm{~K}\), the equilibrium constants for the reactions (1) \(2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) and \((2)\) \(2 \mathrm{CO}_{2}(\mathrm{~g})=2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) are \(1.6 \times 10^{-11}\) and \(1.3 \times 10^{-10}\), respectively. (a) What is the equilibrium constant for the reaction (3) \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})=\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g})\) at that temperature? (b) Show that the manner in which equilibrium constants are calculated is consistent with the manner in which the \(\Delta G_{1}^{\circ}\) values are calculated when combining two or more equarions by determining \(\Delta G_{e}{ }^{\circ}\) for \((1)\) and \((2)\) and using those values to calculare \(\Delta G,{ }^{\circ}\) and \(K_{3}\) for reaction (3).

A reaction mixture consisting of \(2.00 \mathrm{~mol} \mathrm{CO}\) and \(3.00 \mathrm{~mol} \mathrm{H}_{2}\) is placed into a \(10.0-\mathrm{L}\) reaction vessel and heated to \(1200 \mathrm{~K}\). At cquilibrium, \(0.478 \mathrm{~mol} \mathrm{CH}_{4}\) was present in the system. Determine the value of \(K_{c}\) for the reaction \(\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{~g})+\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).

Write the equilibrium expressions \(K_{c}\) for the following reactions. (a) \(\mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{COCl}(\mathrm{g})+\mathrm{Cl}(\mathrm{g})\) (b) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HBr}(\mathrm{g})\) (c) \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

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