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Two different proteins \(X\) and \(Y\) are dissolved in aqueous solution at \(37^{\circ} \mathrm{C}\) . The proteins bind in a \(1 : 1\) ratio to form \(X Y . A\) solution that is initially 1.00 \(\mathrm{mM}\) in each protein is allowed to reach equilibrium. At equilibrium, 0.20 \(\mathrm{mM}\) of free \(\mathrm{X}\) and 0.20 \(\mathrm{mM}\) of free Y remain. What is \(K_{c}\) for the reaction?

Short Answer

Expert verified
The equilibrium constant, \(K_c\), for the reaction \(X + Y \rightleftharpoons XY\) can be determined using the given initial and equilibrium concentrations of X and Y. From the ICE table, we find that the change in concentration (x) is 0.80 mM and the equilibrium concentration of XY is also 0.80 mM. Using the formula \(K_c = \frac{[XY]}{[X][Y]}\) and substituting the equilibrium concentrations, we find that \(K_c = \frac{0.80}{(0.20)(0.20)} = 20\). Therefore, the equilibrium constant for this reaction is 20.

Step by step solution

01

Write the balanced chemical equation

Firstly, we need to write the balanced chemical equation for the reaction. Since the proteins X and Y bind in a 1:1 ratio to form XY, we can write the chemical equation as: \(X + Y \rightleftharpoons XY\)
02

Set up the ICE table

To find the equilibrium constant (\(K_c\)), we need to know the equilibrium concentrations of all the species involved in the reaction. We are given the initial and equilibrium concentrations of X and Y. We can set up an Initial, Change, and Equilibrium (ICE) table to represent the changes in concentrations during the reaction. | | X | Y | XY | |--------|------|------|------| | Initial| 1.00 | 1.00 | 0 | | Change | -x | -x | +x | | Equilibrium | 0.20 | 0.20 | ? |
03

Determine the change in concentration

From the ICE table, we can determine the change in concentration (x) for each species using the given equilibrium concentrations of X and Y: \(x = 1.00 - 0.20 = 0.80\)
04

Find the equilibrium concentration of XY

Now that we have found the change in concentration (x), we can determine the equilibrium concentration of XY: Initial concentration of XY + Change in concentration = Equilibrium concentration of XY \(0 + 0.80 = 0.80\,\text{mM}\)
05

Calculate the equilibrium constant, \(K_c\)

The equilibrium constant, \(K_c\), can be calculated using the equilibrium concentrations of the species in the chemical equation: \(K_c = \frac{[XY]}{[X][Y]}\) Substitute the equilibrium concentrations: \(K_c = \frac{0.80}{(0.20)(0.20)}\) Calculate the value of \(K_c\): \(K_c = \frac{0.80}{0.04} = 20\) Thus, the equilibrium constant for the reaction is 20.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Protein Binding
In the context of biochemistry and molecular biology, protein binding is a crucial concept. It involves the interaction between two or more proteins or between a protein and other molecules. This is particularly important because it can affect protein function, structure, and overall biological activity. When proteins X and Y bind to form a complex, XY, they are effectively conducting a form of communication necessary for executing biological functions.

Protein interactions like this often occur in a 1:1 ratio, as it happens in our example exercise. In this scenario, equal concentrations of proteins X and Y combine to form a single unit of XY. Understanding these interactions helps scientists learn how proteins regulate processes such as metabolism or signal transduction in cells.

Why is protein binding significant? It influences:
  • Enzyme activity and thus metabolic pathways
  • The regulation of biological processes such as gene expression
  • Cellular signaling and communication
These interactions are reversible, meaning that under certain conditions, the binding relationship can be disrupted to form free proteins again, which is vital in addressing changing cellular needs.
Chemical Equilibrium
Chemical equilibrium represents a state in which the concentrations of reactants and products remain constant over time because the rate of the forward reaction equals the rate of the backward reaction. In the case of the binding of proteins X and Y, once equilibrium is reached, the amount of free proteins stops changing because the formation and dissociation of XY occur at the same rate.

Reaching chemical equilibrium is essential because it enables biological systems to maintain homeostasis. This is the balance within biological systems that ensures optimal operative conditions.

Several factors can affect chemical equilibrium, including:
  • Temperature: As demonstrated in the exercise, temperature can influence equilibration because molecules have more kinetic energy at higher temperatures, potentially altering reaction rates.
  • Concentration: Changes in concentrations of reactants or products can shift the equilibrium, a principle known as Le Chatelier's principle.
Understanding chemical equilibrium involves calculating the equilibrium constant, (\( K_c \)), which quantifies the ratio of products to reactants at equilibrium, revealing how far a reaction proceeds in the forward direction under set conditions.
ICE Table
An ICE table is a systematic way of tracking the concentrations of reactants and products over the course of a chemical reaction to find equilibrium concentrations. The acronym 'ICE' stands for Initial, Change, and Equilibrium, reflecting three stages of this process.

Let's break down how an ICE table works:
  • **Initial:** Begin with the initial concentrations or amounts of each reactant and product. In our example, the initial concentrations of proteins X and Y are both 1.00 mM.
  • **Change:** As the reaction progresses, the change in concentration (\(x\)) is observed. In this context, if a certain amount is consumed, we denote it with a negative sign, while production is marked as positive.
  • **Equilibrium:** Finally, you'll have the equilibrium concentrations, reached after accounting for the changes that have occurred. For the protein binding, the equilibrium concentrations for both X and Y become 0.20 mM, meaning the change (\(x\)) was 0.80 mM.
Using this ICE table method allows you to carefully assess chemical reactions and determine crucial values, such as \( K_c \), by organizing data in a clear framework. This visual aid is especially useful for complex reactions and when dealing with multiple components interacting simultaneously.

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Most popular questions from this chapter

(a) Is the dissociation of fluorine molecules into atomic fluorine, \(F_{2}(g) \rightleftharpoons 2 \mathrm{F}(g)\) an exothermic or endothermic process? (b) If the temperature is raised by \(100 \mathrm{K},\) does the equilibrium constant for this reaction increase or decrease? (c) If the temperature is raised by 100 \(\mathrm{K}\) , does the forward rate constant \(k_{f}\) increase by a larger or smaller amount than the reverse rate constant \(k_{r} ?\)

Suppose that the gas-phase reactions A \(\longrightarrow \mathrm{B}\) and \(\mathrm{B} \longrightarrow\) A are both elementary processes with rate constants of \(4.7 \times 10^{-3} \mathrm{s}^{-1}\) and \(5.8 \times 10^{-1} \mathrm{s}^{-1}\) , respectively. (a) What is the value of the equilibrium constant for the equilibrium \(A(g) \rightleftharpoons B(g) ?(\mathbf{b})\) Which is greater at equilibrium, the partial pressure of A or the partial pressure of \(B ?\)

Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps: $$\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \operatorname{COCl}(g)+\mathrm{Cl}(g)$$ At \(25^{\circ} \mathrm{C},\) the rate constants for the forward and reverse reactions are \(1.4 \times 10^{-28} M^{-1} \mathrm{s}^{-1}\) and \(9.3 \times 10^{10} M^{-1} \mathrm{s}^{-1}\) respectively. (a) What is the value for the equilibrium constant at \(25^{\circ} \mathrm{C} ?\) (b) Are reactants or products more plentiful at equilibrium?

The reaction of an organic acid with an alcohol, in organic solvent, to produce an ester and water is commonly done in the pharmaceutical industry. This reaction is catalyzed by strong acid (usually \(\mathrm{H}_{2} \mathrm{SO}_{4} ) .\) A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water: $$\begin{aligned} \mathrm{CH}_{3} \mathrm{COOH}(s o l v)+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(s o l v) & \rightleftharpoons \\ \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(\mathrm{solv}) &+\mathrm{H}_{2} \mathrm{O}(\text {solv}) \end{aligned}$$ where \(^{a}(s o l v)^{\prime \prime}\) indicates that all reactants and products are in solution but not an aqueous solution. The equilibrium constant for this reaction at \(55^{\circ} \mathrm{C}\) is 6.68 . A pharmaceutical chemist makes up 15.0 \(\mathrm{L}\) of a solution that is initially 0.275 \(\mathrm{M}\) in acetic acid and 3.85\(M\) in ethanol. At equilibrium, how many grams of ethyl acetate are formed?

How do the following changes affect the value of the equilibrium constant for a gas-phase exothermic reaction: (a) removal of a reactant, (b) removal of a product, (c) decrease in the volume, (d) decrease in the temperature, (e) addition of a catalyst?

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