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How do the following changes affect the value of the \(K_{P}\) for a gas-phase endothermic reaction: (a) increase in the total pressure by adding a noble gas, \((\mathbf{b})\) addition of a reactant, \((\mathbf{c})\) increase in the temperature (d) increase in the volume, \((\mathbf{e})\) decrease in the temperature?

Short Answer

Expert verified
The effect of the changes on the value of the \(K_P\) for a gas-phase endothermic reaction are: (a) Adding a noble gas does not affect the value of \(K_P\) as it doesn't interact with the reactants. (b) Addition of a reactant doesn't change the value of \(K_P\) as the equilibrium is re-established. (c) An increase in temperature causes the \(K_P\) value to increase as the reaction shifts towards the product side. (d) An increase in volume causes the \(K_P\) value to increase or decrease, depending on whether there are more moles of gas molecules on the product side or the reactant side. (e) A decrease in temperature causes the \(K_P\) value to decrease as the reaction shifts towards the reactant side.

Step by step solution

01

Use Le Chatelier's Principle to Analyze the Effect of Pressure Increase due to Noble Gas

For an endothermic reaction, if we increase the total pressure by adding a noble gas, this won't result in new collisions because noble gases don't interact with the reactants. As a result, the change in pressure won't affect the equilibrium constant \(K_P\). Therefore, there is no change in the value of \(K_P\) in this case.
02

Analyze the Effect of Reactant Addition on the \(K_P\) Value

The addition of a reactant will shift the reaction towards the product according to Le Chatelier's Principle. This will temporarily affect the concentration of reactants and products but at equilibrium, the reaction comes back and the \(K_P\) value remains the same. So, in this case, there is no change in the value of \(K_P\).
03

Determine the Effect of Temperature Increase on the \(K_P\) Value

Increasing the temperature for an endothermic reaction will lead to the reaction absorbing more heat. Le Chatelier's Principle states that the reaction will shift towards the side that can use that extra heat, which means it will shift towards the product side. As the reaction shifts in the direction of the products, the value of \(K_P\) will increase.
04

Explain the Effect of Volume Increase on the \(K_P\) Value

When the volume of the container for a gas-phase endothermic reaction is increased, the pressure decreases. According to Le Chatelier's Principle, the reaction would shift towards the side having larger moles of gaseous components to increase the pressure. Therefore, if the number of moles of gas molecules is larger on the product side, then \(K_P\) will increase. However, if the moles of gas molecules are larger on the reactant side, then \(K_P\) will decrease.
05

Discuss the Effect of Temperature Decrease on the \(K_P\) Value

Decreasing the temperature for an endothermic reaction will result in the reaction absorbing less heat. According to Le Chatelier's Principle, the reaction will shift towards the side that can produce more heat, which means it will shift towards the reactant side. As the reaction shifts in the direction of the reactants, the value of \(K_P\) will decrease.

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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, often denoted as \( K \), is a crucial concept in chemistry. It quantifies the concentrations of reactants and products during a chemical reaction at equilibrium. For reactions involving gases, the equilibrium constant is represented as \( K_P \), referring to partial pressures.

It essentially tells us how far a reaction will proceed before reaching equilibrium. The value of \( K \) is a unique indicator for each reaction at a given temperature:
  • If \( K \) is large, the equilibrium position favors products, indicating the reaction is largely completed.
  • If \( K \) is small, the equilibrium position favors reactants.
Changing conditions such as concentration, pressure, or temperature can affect the equilibrium position, but not always the value of \( K \).

However, it is important to understand that many external changes, like adding a reactant or noble gas, influence the equilibrium state temporarily. This results in an imbalance which Le Chatelier's Principle seeks to remedy until new equilibrium is reached. Yet, such changes do not influence the \( K \) value itself unless the temperature is altered.
Endothermic Reaction
An endothermic reaction is one that absorbs heat from its surroundings, elevating energy levels in the process. A classic trait of these reactions is their positive enthalpy change, denoted as \( \Delta H \).

During such reactions, energy is utilized to break bonds in the reactants, requiring more energy than is released when the products form. As a result, endothermic reactions feel cool to the touch.

Here are some key characteristics:
  • An increase in temperature causes the reaction to absorb additional heat, driving the equilibrium towards the products.
  • In conditions where temperature decreases, the reaction releases less heat and the equilibrium shifts towards the reactants.
Understanding how temperature affects these reactions is fundamental to grasping their behavior and the changes in equilibrium constants, especially under the application of Le Chatelier's Principle.
Gas-Phase Reaction
Gas-phase reactions are chemical processes that involve gaseous reactants and products. These reactions are distinctive due to their sensitivity to changes in pressure, volume, and temperature.

In these reactions, knowing the number of moles of gases involved is crucial since:
  • An increase in pressure or decrease in volume will typically favor the side with fewer gas molecules, shifting the equilibrium accordingly.
  • An increase in volume provides more space, so the equilibrium may shift to the side with more gas molecules.
For gas-phase endothermic reactions, changes in volume and pressure can influence the equilibrium, but only temperature changes affect the equilibrium constant directly. Understanding these dynamics assists in predicting the probable outcomes and shifts in equilibria for gas-phase reactions as dictated by Le Chatelier's Principle.

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Most popular questions from this chapter

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 \(K_{p \cdot}(\mathbf{b})\) 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?

Consider the equilibrium $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) $$ Calculate the equilibrium constant \(K_{p}\) for this reaction, given the following information (at \(298 \mathrm{~K}\) ): $$ \begin{array}{l} 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) \quad K_{c}=2.0 \\ 2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \quad K_{c}=2.1 \times 10^{30} \end{array} $$

At \(1200 \mathrm{~K}\), the approximate temperature of automobile exhaust gases (Figure 15.15 ), \(K_{p}\) for the reaction $$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$$ is about \(1 \times 10^{-11}\). Assuming that the exhaust gas (total pressure \(101.3 \mathrm{kPa}\) ) 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 \(\mathrm{CO}_{2}\) reaction? Based on your conclusion, would the CO concentration in the exhaust be decreased or increased by a catalyst that speeds up the \(\mathrm{CO}_{2}\) reaction? Recall that at a fixed pressure and temperature, volume \(\%=\mathrm{mol} \%\).

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be made by the reaction of \(\mathrm{CO}\) with \(\mathrm{H}_{2}:\) $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)$$ (a) Use thermochemical data in Appendix \(\mathrm{C}\) to calculate \(\Delta H^{\circ}\) for this reaction. (b) To maximize the equilibrium yield of methanol, would you use a high or low temperature? (c) To maximize the equilibrium yield of methanol, would you use a high or low pressure?

Bromine and hydrogen react in the gas phase to form hydrogen bromide: \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) .\) The reaction enthalpy is \(\Delta H^{\circ}=-6 \mathrm{~kJ} .\) (a) To increase the equilibrium yield of hydrogen bromide would you use high or low temperature? (b) Could you increase the equilibrium yield of hydrogen bromide by controlling the pressure of this reaction? If so, would high or low pressure favor formation of \(\mathrm{HBr}(g) ?\)

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