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Chapter 12: Problem 6

Consider the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g) .\) A friend asks the following: "I know we have been told that if a mixture of \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) is at equilibrium and more of \(\mathrm{A}\) is added, more \(\mathrm{C}\) and \(\mathrm{D}\) will form. But how can more \(\mathrm{C}\) and \(\mathrm{D}\) form if we do not add more \(\mathrm{B} ?^{\prime \prime}\) What do you tell your friend?

Short Answer

Expert verified
Adding more reactant A to the system at equilibrium according to Le Châtelier's principle will shift the reaction in the direction that favors the formation of products C and D. As a result, the reaction will consume more A and B to restore equilibrium, leading to higher concentrations of C and D, even without adding more reactant B.

Step by step solution

01

Explain Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction (formation of products) is equal to the rate of the reverse reaction (breakdown of products). In other words, the concentrations of reactants and products no longer change, even though the reaction is still happening at the molecular level, just at equal rates in both directions.
02

Describe Le Châtelier's Principle

Le Châtelier's Principle states that if a system at equilibrium is subjected to a change in conditions, such as concentration, temperature, or pressure, the system will adjust itself to counteract the change and restore a new equilibrium. Essentially, the reaction will shift in the direction that opposes the change.
03

Adding More Reactant A

Now, let's consider the case where we add more of reactant A to the system at equilibrium. According to Le Châtelier's principle, the reaction will shift in a direction to counteract the increase in A's concentration. In this case, the forward reaction (A + B ⇌ C + D) will increase its rate in order to consume the added A and restore equilibrium.
04

No Need to Add More Reactant B

Because only A has been added, the concentration of B initially remains the same. However, as the forward reaction increases its rate and consumes more A and B to form C and D, the concentration of B will start to decrease as well. As a result, the equilibrium will shift to accommodate the decrease in B by further favoring product formation and consume more A to restore the balance. This process will continue until a new equilibrium is reached, with higher concentrations of C and D than before. So, even though we did not add more B to the system, the reaction's shift to restore equilibrium will lead to the formation of more products C and D, as predicted by Le Châtelier's principle.

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Most popular questions from this chapter

At a given temperature, \(K=1.3 \times 10^{-2}\) for the reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\).Calculate values of \(K\) for the following reactions at this temperature. a. \(\frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\). b. \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) c. \(\mathrm{NH}_{3}(g) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)\). d. \(2 \mathrm{N}_{2}(g)+6 \mathrm{H}_{2}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)\).

A sample of \(S_{8}(g)\) is placed in an otherwise empty rigid container at \(1325 \mathrm{K}\) at an initial pressure of 1.00 atm, where it decomposes to \(\mathrm{S}_{2}(g)\) by the reaction $$\mathrm{S}_{8}(g) \rightleftharpoons 4 \mathrm{S}_{2}(g)$$. At equilibrium, the partial pressure of \(S_{8}\) is 0.25 atm. Calculate \(K_{\mathrm{p}}\) for this reaction at \(1325 \mathrm{K}\)

Consider an equilibrium mixture of four chemicals \((\mathrm{A}, \mathrm{B}, \mathrm{C}\) and \(\mathrm{D},\) all gases) reacting in a closed flask according to the equation:$$\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)$$. a. You add more \(A\) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. b. You have the original setup at equilibrium, and you add more \(\mathrm{D}\) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

"Old-fashioned "smelling salts" consist of ammonium carbon=ate, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} .\) The reaction for the decomposition of ammomium carbonate $$\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$,is endothermic. Would the smell of ammonia increase or decrease as the temperature is increased?

Nitric oxide and bromine at initial partial pressures of 98.4 and 41.3 torr, respectively, were allowed to react at \(300 .\) K. At equilibrium the total pressure was 110.5 torr. The reaction is $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ a. Calculate the value of \(K_{\mathrm{p}}\). b. What would be the partial pressures of all species if NO and \(\mathrm{Br}_{2},\) both at an initial partial pressure of 0.30 atm, were allowed to come to equilibrium at this temperature?
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