Cyclohexane \((C)\) and methylcyclopentane \((M)\) are isomers with the chemical formula \(\mathrm{C}_{6} \mathrm{H}_{12}\). The equilibrium constant for the rearrangement \(\mathrm{C} \rightleftharpoons \mathrm{M}\) in solution is \(0.140\) at \(25^{\circ} \mathrm{C}\). (a) \(\mathrm{A}\) solution of \(0.0200 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) cyclohexane and \(0.100 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) methylcyclopentane is prepared. Is the system at equilibrium? If not, will it will form more reactants or more products? (b) What are the concentrations of cyclohexane and methylcyclohexane at equilibrium? (c) If the temperature is raised to \(50 .{ }^{\circ} \mathrm{C}\), the concentration of cyclohexane becomes \(0.100 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) when equilibrium is re-established. Calculate the new equilibrium constant. (d) Is the reaction exothermic or endothermic at \(25^{\circ} \mathrm{C}\) ? Explain your conclusion.

Short Answer

Expert verified
The system is not at equilibrium initially, and at 25°C, more reactants will form. At equilibrium, [C] = 0.0277 mol/L and [M] = 0.0923 mol/L. The new equilibrium constant at 50°C is 0.800. The reaction is endothermic at 25°C since the equilibrium constant increases with temperature.

Step by step solution

01

Determining if the system is at equilibrium

To determine if the system is at equilibrium, calculate the reaction quotient, Q, using the initial concentrations and compare it to the equilibrium constant, K. The reaction quotient Q is given by the expression Q = [M]/[C]. Using the initial concentrations, Q = 0.100 / 0.0200 = 5. The equilibrium constant K is 0.140. Since Q > K, the system is not at equilibrium and the reaction will proceed to the left, forming more reactants (Cyclohexane).
02

Calculating the concentrations at equilibrium

Let x be the change in concentration of Cyclohexane (C) that shifts to Methylcyclopentane (M). At equilibrium, [C] = 0.0200 + x and [M] = 0.100 - x. Plugging these values into the equilibrium expression 0.140 = ([M])/([C]), we get 0.140 = (0.100 - x)/(0.0200 + x). Solving for x will provide the changes in concentrations from the initial values to the equilibrium values. After solving, x is found to be 0.00768 mol/L. Therefore, at equilibrium, [C] = 0.0277 mol/L and [M] = 0.0923 mol/L.
03

Calculating the new equilibrium constant at 50°C

The new equilibrium concentration of Cyclohexane (C) is given to be 0.100 mol/L. Since the new temperature affects the equilibrium position, we assume the new concentration of Methylcyclopentane (M) would be the initial [M] minus the change in [C] (0.100 - 0.0200), which is 0.0800 mol/L. The new equilibrium constant K' can be calculated using K' = [M]/[C]. Thus, K' = 0.0800 / 0.100 = 0.800.
04

Determining if the reaction is exothermic or endothermic

Since the equilibrium constant increases with temperature from 0.140 at 25°C to 0.800 at 50°C, the forward reaction is favored at higher temperatures. Therefore, the reaction must absorb energy (heat) in proceeding from reactants to products, indicating it is endothermic.

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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, represented by the symbol \(K\), is a critical aspect of chemical equilibrium. It reflects the ratio of the concentration of products to reactants at equilibrium for a reversible chemical reaction. In essence, the value of \(K\) determines the extent of the reaction; a larger \(K\) suggests the equilibrium favors the formation of products, whereas a smaller \(K\) indicates a predilection towards the reactants.

Taking the textbook exercise as an example, the equilibrium constant for the cyclohexane and methylcyclopentane rearrangement is \(0.140\) at \(25^\circ C\). This value is used to assess whether the provided concentrations of reactants and products are at equilibrium or if the reaction will shift to balance itself. By comparing \(K\) with the calculated reaction quotient \(Q\), which is based on the current concentrations, one can anticipate the direction in which the reaction will proceed to achieve equilibrium.
Reaction Quotient
The reaction quotient, \(Q\), serves as a 'snapshot' of a reaction's composition at any moment and can be compared against the equilibrium constant, \(K\), to predict the direction of shift towards equilibrium. It is calculated using the same formula as \(K\), but with the current concentrations of reactants and products rather than those at equilibrium.

In our cyclohexane and methylcyclopentane scenario, by calculating \(Q = \frac{[M]}{[C]} = \frac{0.100}{0.0200} = 5\) and observing that \(Q > K\), we deduce that the reaction mixture is not at equilibrium and will move towards the formation of more cyclohexane, the reactant, to lower the value of \(Q\) to match \(K\).
Le Chatelier's Principle
Le Chatelier's principle provides insight on how a system at equilibrium responds to external stresses, such as changes in concentration, pressure, or temperature. It posits that a system will adjust to counteract the effect of the disturbance and re-establish equilibrium. This principle is fundamental when predicting the outcome of perturbations to the equilibrium state.

For instance, increasing the temperature of an endothermic reaction will shift the equilibrium to the right towards the products to absorb the extra heat, as seen in the cyclohexane conversion in our exercise when the temperature is raised from \(25^\circ C\) to \(50^\circ C\). Conversely, cooling an exothermic reaction will prompt the equilibrium to shift to the right to release heat and restore balance.
Endothermic and Exothermic Reactions
Chemical reactions can either release energy (exothermic) or absorb it (endothermic), with these energy changes influencing equilibrium. An exothermic reaction, which releases energy in the form of heat, will have an increased rate of the reverse reaction when heated, resulting in a decrease in the value of the equilibrium constant \(K\). On the other hand, an endothermic reaction absorbs energy, and as such, an increase in temperature will cause an increase in the value of \(K\) and the formation of more products.

Through our example—wherein the equilibrium constant rises from 0.140 at \(25^\circ C\) to 0.800 at \(50^\circ C\)—we can deduce the reaction is endothermic. The system absorbs additional heat upon heating, favoring the formation of products and reflecting a higher \(K\) value at the elevated temperature.

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Most popular questions from this chapter

Analysis of a reaction mixture showed that it had the composition \(0.624 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}, 0.315 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\), and \(0.222 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{NH}_{3}\) at \(800 . \mathrm{K}\), at which temperature \(K_{c}=0.278\) for \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}\) (g). (a) Calculate the reaction quotient \(Q_{c}\). (b) Is the reaction mixture at equilibrium? (c) If not, is there a tendency to form more reactants or more products?

At \(500 .{ }^{\circ} \mathrm{C}, K_{c}=0.061\) for \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\). If analysis shows that the composition is \(3.00 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}\), \(2.00 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\), and \(0.500 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{NH}_{3}\), is the reaction at equilibrium? If not, in which direction does the reaction tend to proceed to reach equilibrium?

Consider the equilibrium \(3 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+\) \(6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\). (a) What happens to the partial pressure of \(\mathrm{NH}_{3}\) when the partial pressure of \(\mathrm{NO}\) is increased? (b) Does the partial pressure of \(\mathrm{O}_{2}\) decrease when the partial pressure of \(\mathrm{NH}_{3}\) is decreased?

Calculate the reaction Gibbs free energy of \(\mathrm{H}_{2}(\mathrm{~g})+\) \(\mathrm{I}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g})\) at 700 . \(\mathrm{K}\) when the partial pressures of \(\mathrm{H}_{2}, \mathrm{I}_{2}\), and HI are \(0.35\) bar, \(0.18\) bar, and \(2.85\) bar, respectively. For this reaction, \(K=54\) at 700 . \(\mathrm{K}\). (b) Indicate whether this reac mixture is likely to form reactants, is likely to form produ is at equilibrium.

The four gases \(\mathrm{NH}_{3}, \mathrm{O}_{2}, \mathrm{NO}\), and \(\mathrm{H}_{2} \mathrm{O}\) are mixed in a reaction vessel and allowed to reach equilibrium in the reaction \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\). Certain changes (see the following table) are then made to this mixture. Considering each change separately, state the effect (increase, decrease, or no change) that the change has on the original equilibrium values of the quantity in the second column (or \(K\), if that is specified). The temperature and volume are constant. \(\begin{array}{ll}\text { Change } & \text { Quantity } \\ \text { (a) add } \mathrm{NO} & \text { amount of } \mathrm{H}_{2} \mathrm{O} \\ \text { (b) add } \mathrm{NO} & \text { amount of } \mathrm{O}_{2} \\ \text { (c) remove } \mathrm{H}_{2} \mathrm{O} & \text { amount of } \mathrm{NO} \\ \text { (d) remove } \mathrm{O}_{2} & \text { amount of } \mathrm{NH}_{3} \\ \text { (e) add } \mathrm{NH}_{3} & \mathrm{~K} \\ \text { (f) remove } \mathrm{NO} & \text { amount of } \mathrm{NH}_{3} \\ \text { (g) add } \mathrm{NH}_{3} & \text { amount of } \mathrm{O}_{2}\end{array}\)

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