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

Short Answer

Expert verified
The equilibrium constant, \(K_c\), for the reaction between proteins X and Y can be calculated using the expression \[ K_c = \frac{[\mathrm{XY}]}{[\mathrm{X]}\times[\mathrm{Y}]}\] Given the initial concentrations and the equilibrium concentrations of X and Y, we find the concentration of XY at equilibrium and substitute the values into the expression. The calculated \(K_c\) for this reaction is \(20\).

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation for the interaction between proteins X and Y can be written as follows: \[ X + Y \rightleftharpoons XY \]
02

Set up the initial and equilibrium concentrations table

We can set up a table to show the initial concentrations of X and Y and their equilibrium concentrations. The initial concentrations are both \(1.00 \mathrm{~m} M\), and at equilibrium, there are \(0.20 \mathrm{~m} M\) of free X and Y remaining. We can then use this information to determine the concentration of XY at equilibrium. \begin{tabular}{c|ccc} & X & Y & XY \\ \hline Initial & 1.00 & 1.00 & 0 \\ Change & -0.80 & -0.80 & +0.80 \\ Equilibrium & 0.20 & 0.20 & 0.80 \\ \end{tabular}
03

Write the expression for \(K_c\)

The expression for the equilibrium constant, \(K_c\), for this reaction can be written as follows: \[ K_c = \frac{[\mathrm{XY}]}{[\mathrm{X]}\times[\mathrm{Y}]} \]
04

Substitute the equilibrium concentrations into the \(K_c\) expression

Now let's plug the equilibrium concentrations from the table above into the \(K_c\) expression: \[ K_c = \frac{0.80}{(0.20) \times (0.20)} \]
05

Calculate the value of \(K_c\)

Finally, we calculate the value of \(K_c\) by solving the equation: \[ K_c = \frac{0.80}{(0.20) \times (0.20)} = \frac{0.80}{0.04} = 20 \] So, the equilibrium constant \(K_c\) for this reaction is \(20\).

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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium is essential for grasping how reactions proceed and the conditions under which they stabilize. It refers to a state in a chemical process where the rates of the forward and reverse reactions are equal, resulting in no further net change in the concentrations of reactants and products over time. At this point, it might seem like the reaction has ceased, but in reality, it continues with reactants converting to products and vice versa at an equal rate.

This balance does not imply that the reactants and products have the same concentration, but rather that their concentrations have reached a point where they remain constant. Key to this concept is the dynamic nature of the equilibrium state; although the concentrations do not change any further, the reaction is still occurring on the molecular level.
Equilibrium Concentrations

The term 'equilibrium concentrations' refers to the amounts of reactants and products present when a chemical reaction has attained equilibrium. These concentrations can be calculated by considering the stoichiometry of the reaction and the initial concentrations of the substances involved. For the example of the proteins X and Y binding to form XY, a table is used to compare initial and equilibrium concentrations to calculate changes in concentration (-0.80 mM for X and Y, and +0.80 mM for XY in the example).

Importance of Stoichiometry

Stoichiometry is pivotal in deducing the changes in concentration as it dictates the proportion in which the reactants combine. By understanding this concept, you can predict how a change in one substance affects the concentration of another.

Initial versus Equilibrium Concentrations

Initial concentrations are the starting amounts of reactants or products prior to any reaction taking place, whereas equilibrium concentrations are measured when the system has stabilized. Recognizing this distinction is crucial in equilibrium calculations as it helps in setting up a proper reaction table and determining the reaction quotient, which leads to finding the equilibrium constant.

Reaction Quotient
The reaction quotient, denoted as Q, is a measure that compares the relative amounts of products and reactants present during a reaction at any point before reaching equilibrium. It has the same form as the equilibrium constant expression but is calculated using the concentrations at any moment in time, not just at equilibrium.

For a general reaction where aA + bB ⇌ cC + dD, the reaction quotient is given by
\[ Q = \frac{[C]^c[D]^d}{[A]^a[B]^b} \] As the reaction proceeds, Q changes until it eventually equals the equilibrium constant (K). If Q < K, the forward reaction is favored, and the system will shift towards producing more products. Conversely, if Q > K, the reaction will proceed in the reverse direction to produce more reactants. This concept is key for predicting the direction of the shift in reactant and product concentrations as a reaction moves towards equilibrium.

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Most popular questions from this chapter

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?

Suppose that the gas-phase reactions \(\mathrm{A} \longrightarrow \mathrm{B}\) and \(\mathrm{B} \longrightarrow \mathrm{A}\) 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 \(\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g) ?\) (b) Which is greater at equilibrium, the partial pressure of \(\mathrm{A}\) or the partial pressure of \(\mathrm{B}\) ? Explain.

At \(2000^{\circ} \mathrm{C}\) the equilibrium constant for the reaction $$2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)$$ is \(K_{c}=2.4 \times 10^{3}\). If the initial concentration of \(\mathrm{NO}\) is \(0.175 \mathrm{M}\) what are the equilibrium concentrations of \(\mathrm{NO}, \mathrm{N}_{2},\) and \(\mathrm{O}_{2} ?\)

At \(900 \mathrm{~K}\) the following reaction has \(K_{p}=0.345\) : $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ In an equilibrium mixture the partial pressures of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are 0.135 atm and 0.455 atm, respectively. What is the equilibrium partial pressure of \(\mathrm{SO}_{3}\) in the mixture?

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen: \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) .\) An equilibrium mixture in a \(2.00-\mathrm{L}\) vessel is found to contain \(0.0406 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{OH},\) \(0.170 \mathrm{~mol} \mathrm{CO},\) and \(0.302 \mathrm{~mol} \mathrm{H}_{2}\) at \(500 \mathrm{~K} .\) Calculate \(K_{c}\) at this temperature.

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