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Chapter 13: Problem 107

Consider the following exothermic reaction at equilibrium: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ Predict how the following changes affect the number of moles of each component of the system after equilibrium is reestablished by completing the table below. Complete the table with the terms increase, decrease, or no change.

Short Answer

Expert verified
In summary, applying Le Chatelier's Principle to the given exothermic reaction and analyzing different disturbances yields the following changes in the number of moles for each case: 1. Increasing moles of N2: - N2: Increase - H2: Decrease - NH3: Increase 2. Decreasing moles of H2: - N2: Increase - H2: Decrease - NH3: Decrease 3. Increasing moles of NH3: - N2: Increase - H2: Increase - NH3: Increase 4. Decreasing moles of NH3: - N2: Decrease - H2: Decrease - NH3: Decrease

Step by step solution

01

Le Chatelier's Principle

Le Chatelier's principle states that when a chemical system at equilibrium is disturbed, the system will adjust itself to counteract the change and restore the equilibrium. In this case, we will apply this principle to predict how the changes in the system will affect the number of moles of each component. We have the following equilibrium reaction: \[\mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\] Now, let's examine each row in the given table. 1. Increasing the moles of N2:
02

Forward reaction

Adding more N2 will shift the equilibrium towards the forward reaction, increasing the formation of NH3. Consequently, the moles of H2 will decrease as they are being consumed in the reaction. - N2: Increase - H2: Decrease - NH3: Increase 2. Decreasing the moles of H2:
03

Reverse reaction

Removing H2 will shift the equilibrium towards the reverse reaction to compensate for the loss, leading to decreased formation of NH3. The backward reaction will produce more N2. - N2: Increase - H2: Decrease - NH3: Decrease 3. Increasing the moles of NH3:
04

Reverse reaction

Adding more NH3 will shift the equilibrium towards the reverse reaction. Consequently, the moles of N2 and H2 will increase as NH3 is being consumed in the backward reaction. - N2: Increase - H2: Increase - NH3: Increase 4. Decreasing the moles of NH3:
05

Forward reaction

Removing NH3 will shift the equilibrium towards the forward reaction to compensate for the loss, leading to increased formation of NH3. The forward reaction will consume more N2 and H2. - N2: Decrease - H2: Decrease - NH3: Decrease These are the predicted changes in the number of moles for each component in the reaction based on Le Chatelier's Principle.

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Most popular questions from this chapter

The synthesis of ammonia gas from nitrogen gas and hydrogen gas represents a classic case in which a knowledge of kinetics and equilibrium was used to make a desired chemical reaction economically feasible. Explain how each of the following conditions helps to maximize the yield of ammonia. a. running the reaction at an elevated temperature b. removing the ammonia from the reaction mixture as it forms c. using a catalyst d. running the reaction at high pressure

At high temperatures, elemental nitrogen and oxygen react with each other to form nitrogen monoxide: $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ Suppose the system is analyzed at a particular temperature, and the equilibrium concentrations are found to be \(\left[\mathrm{N}_{2}\right]=\) \(0.041 M,\left[\mathrm{O}_{2}\right]=0.0078 M,\) and $[\mathrm{NO}]=4.7 \times 10^{-4} M .\( Calculate the value of \)K$ for the reaction.

The reaction $$\mathrm{NH}_{4} \mathrm{SH}(s) \leftrightharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g)$$ has \(K_{\mathrm{p}}=0.10\) at \(27^{\circ} \mathrm{C} .\) What is the minimum amount of \(\mathrm{NH}_{4} \mathrm{SH}\) that must be present for this reaction to be at equilibrium in a 10.0 \(\mathrm{-L}\) container?

For the reaction $$\mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{HS}(s)$$ \(K=400\) . at \(35.0^{\circ} \mathrm{C} .\) If 2.00 moles each of \(\mathrm{NH}_{3}, \mathrm{H}_{2} \mathrm{S},\) and $\mathrm{NH}_{4} \mathrm{HS}\( are placed in a \)5.00-\mathrm{L}$ vessel, what mass of \(\mathrm{NH}_{4} \mathrm{HS}\) will be present at equilibrium? What is the pressure of \(\mathrm{H}_{2} \mathrm{S}\) at equilibrium?

The following equilibrium pressures were observed at a certain temperature for the reaction $$\begin{array}{c}{\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)} \\ {P_{\mathrm{NH}_{3}}=3.1 \times 10^{-2} \mathrm{atm}} \\\ {P_{\mathrm{N}_{2}}=8.5 \times 10^{-1} \mathrm{atm}} \\\ {P_{\mathrm{H}_{2}}=3.1 \times 10^{-3} \mathrm{atm}}\end{array}$$ Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature. If $P_{\mathrm{N}_{2}}=0.525 \mathrm{atm}, P_{\mathrm{NH}_{3}}=0.0167 \mathrm{atm},\( and \)P_{\mathrm{H}_{2}}=0.00761 \mathrm{atm},$ does this represent a system at equilibrium?
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