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Predict the effect of increasing the container volume on the amounts of each reactant and product in the following reactions: (a) \(\mathrm{CH}_{3} \mathrm{OH}(l) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)\) (b) \(\mathrm{CH}_{4}(g)+\mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{HCN}(g)+3 \mathrm{H}_{2}(g)\)

Short Answer

Expert verified
Increasing volume favors the gaseous products in (a) and the side with more moles of gas in (b).

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.
02

- Analyze Reaction (a)

Consider the reaction \(\text{CH}_3\text{OH}(l) \rightleftharpoons \text{CH}_3\text{OH}(g)\). When the volume of the container is increased, the pressure decreases. According to Le Chatelier's Principle, the system will shift to increase the pressure, which means it will shift towards the side with more moles of gas. Here, the gaseous side \( \text{CH}_3\text{OH}(g) \) will be favored. Consequently, the amount of \( \text{CH}_3\text{OH}(l) \) will decrease while the amount of \( \text{CH}_3\text{OH}(g) \) will increase.
03

- Analyze Reaction (b)

Consider the reaction \(\text{CH}_4(g) + \text{NH}_3(g) \rightleftharpoons \text{HCN}(g) + 3 \text{H}_2(g)\). Increasing the container volume will decrease the pressure. The system will shift toward the side with more moles of gas to increase the pressure. Here, the right side has a total of 4 moles of gas compared to 2 moles on the left side. Therefore, the equilibrium will shift towards the right, increasing the amounts of \( \text{HCN}(g) \) and \( \text{H}_2}(g) \) and decreasing the amounts of \( \text{CH}_4(g)\) and \( \text{NH}_3(g)\).

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Key Concepts

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

equilibrium shift
When a chemical reaction is at equilibrium, it means the rate of the forward reaction equals the rate of the reverse reaction. This balance can be tipped in favor of the forward or reverse reaction by changing the conditions. This is known as an equilibrium shift. Le Chatelier's Principle is a helpful guide to predict the direction of this shift when conditions such as pressure, temperature, or concentration are altered. For example, in reaction (a) \(\text{CH}_3\text{OH}(l) \rightleftharpoons \text{CH}_3\text{OH}(g)\), if the volume of the container is increased, the equilibrium will shift towards the side with more moles of gas, which is the gaseous \(\text{CH}_3\text{OH}(g)\) side.
pressure change
Pressure changes can significantly influence the position of equilibrium in a reaction involving gases. According to Le Chatelier's Principle, if the pressure of a system at equilibrium is decreased, the system will shift to increase the pressure again. This usually means moving towards the side with more moles of gas. Conversely, if the pressure is increased, the system will shift towards the side with fewer moles of gas. In reaction (b) \(\text{CH}_4(g) + \text{NH}_3(g) \rightleftharpoons \text{HCN}(g) + 3 \text{H}_2(g)\), increasing the volume of the container (which decreases the pressure) causes the equilibrium to shift to the right, where there are more moles of gas.
reaction dynamics
Reaction dynamics involve understanding how different variables affect the speed and direction of a reaction. Le Chatelier's Principle is key here, because it helps predict how a system will respond to changes and return to equilibrium. By knowing whether a system will shift left or right, you can predict the concentrations of reactants and products. For instance, in our example reaction (b), knowing that increasing the container's volume decreases pressure allows you to predict an increase in the amounts of \(\text{HCN}(g)\) and \(\text{H}_2(g)\).
moles of gas
In a gaseous reaction, the number of moles of gas on each side of the equation plays a crucial role in determining how the equilibrium shifts in response to pressure changes. In reaction (a) \(\text{CH}_3\text{OH}(l) \rightleftharpoons \text{CH}_3\text{OH}(g)\), the equilibrium will shift towards the side with more gas moles (gaseous \(\text{CH}_3\text{OH}(g)\)) when the container volume is increased. Similarly, in reaction (b) \(\text{CH}_4(g) + \text{NH}_3(g) \rightleftharpoons \text{HCN}(g) + 3 \text{H}_2(g)\), the right side has 4 moles of gas compared to 2 moles on the left side, prompting the shift towards the right when the volume is increased.

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Most popular questions from this chapter

What is implied by the word constant in the term equilibrium constant? Give two reaction parameters that can be changed without changing the value of an equilibrium constant.

Sodium bicarbonate undergoes thermal decomposition according to the reaction $$ 2 \mathrm{NaHCO}_{3}(s) \Longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ How does the equilibrium position shift as a result of each of the following disturbances? (a) 0.20 atm of argon gas is added. (b) \(\mathrm{NaHCO}_{3}(s)\) is added. (c) \(\mathrm{Mg}\left(\mathrm{ClO}_{4}\right)_{2}(s)\) is added as a drying agent to remove \(\mathrm{H}_{2} \mathrm{O}\). (d) Dry ice is added at constant \(T\)

Even at high \(T,\) the formation of \(\mathrm{NO}\) is not favored: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \quad K_{\mathrm{c}}=4.10 \times 10^{-4} \text {at } 2000^{\circ} \mathrm{C} $$ What is [NO] when a mixture of \(0.20 \mathrm{~mol}\) of \(\mathrm{N}_{2}(g)\) and \(0.15 \mathrm{~mol}\) of \(\mathrm{O}_{2}(g)\) reaches equilibrium in a \(1.0-\mathrm{L}\) container at \(2000^{\circ} \mathrm{C} ?\)

A key step in the extraction of iron from its ore is \(\mathrm{FeO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) \quad K_{\mathrm{p}}=0.403\) at \(1000^{\circ} \mathrm{C}\) This step occurs in the \(700^{\circ} \mathrm{C}\) to \(1200^{\circ} \mathrm{C}\) zone within a blast furnace. What are the equilibrium partial pressures of \(\mathrm{CO}(g)\) and \(\mathrm{CO}_{2}(g)\) when \(1.00 \mathrm{~atm}\) of \(\mathrm{CO}(g)\) and excess \(\mathrm{FeO}(s)\) react in a sealed container at \(1000^{\circ} \mathrm{C} ?\)

Le Châtelier's principle is related ultimately to the rates of the forward and reverse steps in a reaction. Explain (a) why an increase in reactant concentration shifts the equilibrium position to the right but does not change \(K ;\) (b) why a decrease in \(V\) shifts the equilibrium position toward fewer moles of gas but does not change \(K ;\) (c) why a rise in \(T\) shifts the equilibrium position of an exothermic reaction toward reactants and changes \(K ;\) and (d) why a rise in temperature of an endothermic reaction from \(T_{1}\) to \(T_{2}\) results in \(K_{2}\) being larger than \(K_{1}\)

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