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For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by increasing the temperature and increasing the volume of the reaction vessel. (a) What can you conclude about the reaction from the influence of temperature on the equilibrium? (b) What can you conclude from the influence of increasing the volume?

Short Answer

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(a) Since increasing the temperature increases the fraction of products in equilibrium, the reaction is endothermic. This is because an increase in temperature favors the forward reaction in endothermic processes. (b) The equilibrium shifts to increase the fraction of products when the volume is increased, implying that the number of moles of gas is greater on the product side than on the reactant side.

Step by step solution

01

(a) Influence of Temperature on Equilibrium

Le Chatelier's principle states that if a change in conditions is applied to a system at equilibrium, the system will respond by adjusting its position to counteract the imposed change. When temperature is increased in an exothermic reaction, the equilibrium constant (K) decreases, and when temperature is increased in an endothermic reaction, K increases. In this case, increasing the temperature increases the fraction of products in equilibrium, which implies that K also increases. This further suggests that the reaction is endothermic, as the increase in temperature favors the forward reaction which leads to more products.
02

(b) Influence of Increasing Volume on Equilibrium

When the volume of the reaction vessel is increased, the pressure is generally reduced. According to Le Chatelier's principle, if pressure is decreased, the equilibrium will shift to the side with a greater number of moles of gas, because this will counteract the pressure reduction. In this case, the equilibrium shifts to increase the fraction of products when the volume is increased. This implies that the number of moles of gas is greater on the product side than on the reactant side. So, from (a) and (b), we can conclude that the given reaction is an endothermic reaction with a greater number of moles of gas in products than reactants.

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Key Concepts

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

Le Chatelier's Principle
Le Chatelier's Principle serves as a guide for predicting how a change in conditions impacts a chemical system at equilibrium. It states that if a dynamic equilibrium system is subjected to a change in concentration, temperature, or pressure, the system will adjust itself to partially counteract the effect of the change, ultimately reaching a new equilibrium position.

For instance, in the provided exercise, increasing the temperature caused more products to be formed, indicating that the system shifted to absorb the extra heat. Think of it like an invisible seesaw; when one side is disturbed, the other side adjusts to bring things back into balance. In essence, it's the chemical world's way of saying 'keep calm and carry on.'
Endothermic Reactions
Endothermic reactions absorb heat from their surroundings. This means that as you add heat to the system, such as increasing the temperature, the reaction shifts to consume that heat, producing more product and fewer reactants.

In our exercise, the observation that increasing temperature increases the production of the reaction's products tells us that heat is being used up in the process to form those products, hence confirming that the reaction is endothermic. Unlike their exothermic counterparts, which release heat, endothermic reactions essentially 'soak up' warmth, making them the 'sponges' of the chemical reaction world.
Equilibrium Constant
The equilibrium constant (K) is a number that expresses the ratio of products to reactants at equilibrium, and it varies with temperature. In an endothermic reaction, increasing the temperature increases K, because heat is a reactant. It's similar to saying that more heat 'tips the scales' in favor of the products for endothermic reactions.

The increase in the fraction of products, as mentioned in the exercise, therefore suggests an increase in K, solidifying the notion that the reaction harnesses heat from its surroundings (heat acts as a reactant) to move forward. The value of K provides a snapshot of the reaction's preferred state at a given temperature—a higher K value favors products, hinting at the reaction's cozy relationship with heat.
Reaction Vessel Volume
Changing the volume of the reaction vessel affects the concentration and pressure of the gaseous components in an equilibrium system. An increase in volume leads to a decrease in pressure, causing the reaction to favor the production of more moles of gas, should it be able to, to balance the change.

In our specific scenario, the fact that increasing the volume also increases the fraction of products suggests that there are more gas molecules among the products compared to the reactants. This is akin to opening up a room to allow more party guests (gas molecules) on one side—the system expands to make everyone comfortable, thereby shifting the equilibrium. Always remember, the reaction will shift towards the side with more gas particles when the 'party room' (volume) gets bigger.

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Most popular questions from this chapter

At \(1200 \mathrm{~K}\), the approximate temperature of automobile exhaust gases (Figure 15.16), \(K_{p}\) for the reaction $$ 2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) $$ is about \(1 \times 10^{-13}\). Assuming that the exhaust gas (total pressure \(1 \mathrm{~atm}\) ) contains \(0.2 \% \mathrm{CO}, 12 \% \mathrm{CO}_{2}\), and \(3 \% \mathrm{O}_{2}\) by volume, is the system at equilibrium with respect to the above reaction? Based on your conclusion, would the CO concentration in the exhaust be decreased or increased by a catalyst that speeds up the reaction above?

The reaction \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\) has \(K_{p}=0.0870\) at \(300^{\circ} \mathrm{C}\). A flask is charged with \(0.50\) atm \(\mathrm{PCl}_{3}, 0.50 \mathrm{~atm} \mathrm{Cl}_{2}\), and \(0.20 \mathrm{~atm} \mathrm{PCl}_{5}\) at this tempera- ture. (a) Use the reaction quotient to determine the direction the reaction must proceed to reach equilibrium. (b) Calculate the equilibrium partial pressures of the gases. (c) What effect will increasing the volume of the system have on the mole fraction of \(\mathrm{C}_{2}\) in the equilibrium mixture? (d) The reaction is exothermic. What effect will increasing the temperature of the system have on the mole fraction of \(\mathrm{Cl}_{2}\) in the equilibrium mixture?

The equilibrium constant \(K_{c}\) for \(C(s)+C O_{2}(g) \rightleftharpoons\) \(2 \mathrm{CO}(g)\) is \(1.9\) at \(1000 \mathrm{~K}\) and \(0.133\) at \(298 \mathrm{~K}\). (a) If excess \(\mathrm{C}\) is allowed to react with \(25.0 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) in a 3.00-L vessel at \(1000 \mathrm{~K}\), how many grams of \(\mathrm{CO}_{2}\) are produced? (b) How many grams of \(C\) are consumed? (c) If a smaller vessel is used for the reaction, will the yield of \(\mathrm{CO}\) be greater or smaller? (d) If the reaction is endothermic, how does increasing the temperature affect the equilibrium constant?

Mercury(I) oxide decomposes into elemental mercury and elemental oxygen: \(2 \mathrm{Hg}_{2} \mathrm{O}(s) \rightleftharpoons 4 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) (a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) Explain why we normally exclude pure solids and liquids from equilibrium-constant expressions.

In Section \(11.5\) we defined the vapor pressure of a liquid in terms of an equilibrium. (a) Write the equation representing the equilibrium between liquid water and water vapor, and the corresponding expression for \(K_{p}\). (b) By using data in Appendix \(\mathrm{B}\), give the value of \(K_{p}\) for this reaction at \(30{ }^{\circ} \mathrm{C}\). (c) What is the value of \(K_{p}\) for any liquid in equilibrium with its vapor at the normal boiling point of the liquid?

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