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Chapter 15: Problem 4

Consider the equilibrium system \(3 \mathrm{~A} \rightleftharpoons \mathrm{B}\). Sketch the change in concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) with time for these situations: (a) Initially only A is present; (b) initially only B is present; (c) initially both \(\mathrm{A}\) and \(\mathrm{B}\) are present (with \(\mathrm{A}\) in higher concentration). In each case, assume that the concentration of \(\mathrm{B}\) is higher than that of \(\mathrm{A}\) at equilibrium.

Short Answer

Expert verified
The given examples show how the concentration of reactants and products change over time in an equilibrium system. In all cases, as the reaction progresses, the concentration of A will decrease as it gets converted into B, until equilibrium is reached where the concentration of B is more than A. The rate of this conversion varies depending on the initial concentrations of A and B.

Step by step solution

01

Analyzing the Initial Concentration of A and B

The given information is that initially, only A is present in the first case. This means that the concentration of B is nil, to begin with. As the reaction proceeds, A will begin to decrease as it gets converted into B.
02

Plotting the Change in Concentration Over Time for Case (a)

As the reaction continues, the concentration of A will decrease and the concentration of B will increase as A gets converted into B. This will continue until the equilibrium is reached, where the concentration of B will become higher than that of A.
03

Analyzing the Initial Concentration of A and B for Case (b)

In this case, it is given that solely B is present initially. So, basically, A is nil. As the reaction proceeds, in the reverse direction since B is present, the concentration of B decreases while that of A increases.
04

Plotting the Change in Concentration Over Time for Case (b)

As the reaction continues, the concentration of B will decrease and the concentration of A will increase. This will continue until we attain equilibrium, where the concentration of B is evidently higher than A.
05

Analyzing the Initial Concentration of A and B for Case (c)

Initially, both A and B are present but the concentration of A is higher. As the reaction proceeds, A will start getting converted into B.
06

Plotting the Change in Concentration Over Time for Case (c)

As the reaction continues, the concentration of A will decrease and the concentration of B will increase until we reach a point where B's concentration is greater than A's. After equilibrium, the rate of conversion of A to B will be equal to the rate of conversion of B to A, thus, their concentrations will remain constant.

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Most popular questions from this chapter

For the synthesis of ammonia $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ the equilibrium constant \(K_{\mathrm{c}}\) at \(375^{\circ} \mathrm{C}\) is \(1.2 .\) Starting with \(\left[\mathrm{H}_{2}\right]_{0}=0.76 \mathrm{M},\left[\mathrm{N}_{2}\right]_{0}=0.60 \mathrm{M},\) and \(\left[\mathrm{NH}_{3}\right]_{0}=0.48 M,\) when this mixture comes to equilibrium, which gases will have increased in concentration and which will have decreased in concentration?

Iodine is sparingly soluble in water but much more so in carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\). The equilibrium constant, also called the partition coefficient, for the distribution of \(\mathrm{I}_{2}\) between these two phases $$ \mathrm{I}_{2}(a q) \rightleftharpoons \mathrm{I}_{2}\left(\mathrm{CCl}_{4}\right) $$ is 83 at \(20^{\circ} \mathrm{C}\). (a) A student adds \(0.030 \mathrm{~L}\) of \(\mathrm{CCl}_{4}\) to \(0.200 \mathrm{~L}\) of an aqueous solution containing \(0.032 \mathrm{~g}\) \(\mathrm{I}_{2} .\) The mixture is shaken and the two phases are then allowed to separate. Calculate the fraction of \(\mathrm{I}_{2}\) remaining in the aqueous phase. (b) The student now repeats the extraction of \(I_{2}\) with another 0.030 \(\mathrm{L}\) of \(\mathrm{CCl}_{4}\). Calculate the fraction of the \(\mathrm{I}_{2}\) from the original solution that remains in the aqueous phase. (c) Compare the result in (b) with a single extraction using \(0.060 \mathrm{~L}\) of \(\mathrm{CCl}_{4}\). Comment on the difference.

In the uncatalyzed reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ at \(100^{\circ} \mathrm{C}\) the pressures of the gases at equilibrium are \(P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.377 \mathrm{~atm}\) and \(P_{\mathrm{NO}_{2}}=1.56 \mathrm{~atm} .\) What would happen to these pressures if a catalyst were present?

The equilibrium constant \(K_{\mathrm{c}}\) for the decomposition of phosgene, \(\mathrm{COCl}_{2}\), is \(4.63 \times 10^{-3}\) at \(527^{\circ} \mathrm{C}\) : $$ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) $$ Calculate the equilibrium partial pressure of all the components, starting with pure phosgene at 0.760 atm.

A quantity of 0.20 mole of carbon dioxide was heated at a certain temperature with an excess of graphite in a closed container until the following equilibrium was reached: $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ Under this condition, the average molar mass of the gases was found to be \(35 \mathrm{~g} / \mathrm{mol}\). (a) Calculate the mole fractions of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\). (b) What is the \(K_{P}\) for the equilibrium if the total pressure was 11 atm? (Hint: The average molar mass is the sum of the products of the mole fraction of each gas and its molar mass.)
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