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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 chemist at a pharmaceutical company is measuring equilibrium constants for reactions in which drug candidate molecules bind to a protein involved in cancer. The drug molecules bind the protein in a \(1 : 1\) ratio to form a drug-protein complex. The protein concentration in aqueous solution at \(25^{\circ} \mathrm{C}\) is \(1.50 \times 10^{-6} \mathrm{M}\) . Drug \(\mathrm{A}\) is introduced into the protein solution at an initial concentration of \(2.00 \times 10^{-6} \mathrm{M}\) Drug \(\mathrm{B}\) is introduced into a separate, identical protein solution at an initial concentration of \(2.00 \times 10^{-6} \mathrm{M}\) . At equilibrium, the drug \(\mathrm{A}-\) protein solution has an \(\mathrm{A}\) -protein complex concentration of \(1.00 \times 10^{-6} \mathrm{M},\) and the drug \(\mathrm{B}\) solution has a B-protein complex concentration of \(1.40 \times 10^{-6} \mathrm{M}\) Calculate the \(K_{c}\) value for the A-protein binding reaction and for the B-protein binding reaction. Assuming that the drug that binds more strongly will be more effective, which drug is the better choice for further research?

Methane, \(\mathrm{CH}_{4},\) reacts with \(\mathrm{I}_{2}\) according to the reaction \(\mathrm{CH}_{4}(g)+\mathrm{I}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{I}(g)+\mathrm{HI}(g) .\) At \(630 \mathrm{K}, K_{p}\) for this reaction is \(2.26 \times 10^{-4} .\) A reaction was set up at 630 \(\mathrm{K}\) with initial partial pressures of methane of 105.1 torr and of 7.96 torr for \(\mathrm{I}_{2}\) . Calculate the pressures, in torr, of all reactants and products at equilibrium.

Consider the reaction \(10_{4}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}(a q) ; K_{c}=3.5 \times 10^{-2} .\) If you start with 25.0 \(\mathrm{mL}\) of a 0.905 \(\mathrm{M}\) solution of \(\mathrm{NaIO}_{4},\) and then dilute it with water to 500.0 \(\mathrm{mL}\) , what is the concentration of \(\mathrm{H}_{4} \mathrm{IO}_{6}\) at equilibrium?

NiO is to be reduced to nickel metal in an industrial process by use of the reaction $$\mathrm{NiO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(s)+\mathrm{CO}_{2}(g)$$ At \(1600 \mathrm{K},\) the equilibrium constant for the reaction is \(K_{p}=6.0 \times 10^{2} .\) If a CO pressure of 150 torr is to be employed in the furnace and total pressure never exceeds 760 torr, will reduction occur?

(a) If \(Q_{c}>K_{c}\) how must the reaction proceed to reach equilibrium? (b) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of \(Q_{c}\) at this point in the reaction?

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