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

A mixture of \(\mathrm{H}_{2}, \mathrm{~S},\) and \(\mathrm{H}_{2} \mathrm{~S}\) is held in a \(1.0-\mathrm{L}\) vessel at \(90{ }^{\circ} \mathrm{C}\) and reacts according to the equation: $$\mathrm{H}_{2}(g)+\mathrm{S}(s) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)$$ At equilibrium the mixture contains \(0.46 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{~S}\) and \(0.40 \mathrm{~g}\) \(\mathrm{H}_{2}\) (a) Write the equilibrium-constant expression for this reaction. (b) What is the value of \(K_{c}\) for the reaction at this temperature? (c) Why can we ignore the amount of \(\mathrm{S}\) when doing the calculation in part (b)?

Consider the hypothetical reaction \(\mathrm{A}(g)+2 \mathrm{~B}(g) \rightleftharpoons\) \(2 \mathrm{C}(g),\) for which \(K_{c}=0.25\) at a certain temperature. A \(1.00-\mathrm{L}\) reaction vessel is loaded with \(1.00 \mathrm{~mol}\) of compound \(\mathrm{C}\), which is allowed to reach equilibrium. Let the variable \(x\) represent the number of \(\mathrm{mol} / \mathrm{L}\) of compound A present at equilibrium. (a) In terms of \(x,\) what are the equilibrium concentrations of compounds \(\mathrm{B}\) and \(\mathrm{C} ?\) (b) What limits must be placed on the value of \(x\) so that all concentrations are positive? (c) By putting the equilibrium concentrations (in terms of \(x\) ) into the equilibrium-constant expression, derive an equation that can be solved for \(x\). (d) The equation from part (c) is a cubic equation (one that has the form \(\left.a x^{3}+b x^{2}+c x+d=0\right)\). In general, cubic equations cannot be solved in closed form. However, you can estimate the solution by plotting the cubic equation in the allowed range of \(x\) that you specified in part (b). The point at which the cubic equation crosses the \(x\) -axis is the solution. (e) From the plot in part (d), estimate the equilibrium concentrations of \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\).

Le Châtelier noted that many industrial processes of his time could be improved by an understanding of chemical equilibria. For example, the reaction of iron oxide with carbon monoxide was used to produce elemental iron and \(\mathrm{CO}_{2}\) according to the reaction $$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \rightleftharpoons 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)$$ Even in Le Châtelier's time, it was noted that a great deal of CO was wasted, expelled through the chimneys over the furnaces. Le Châtelier wrote, "Because this incomplete reaction was thought to be due to an insufficiently prolonged contact between carbon monoxide and the iron ore [oxide], the dimensions of the furnaces have been increased. In England they have been made as high as thirty meters. But the proportion of carbon monoxide escaping has not diminished, thus demonstrating, by an experiment costing several hundred thousand francs, that the reduction of iron oxide by carbon monoxide is a limited reaction. Acquaintance with the laws of chemical equilibrium would have permitted the same conclusion to be reached more rapidly and far more economically." What does this anecdote tell us about the equilibrium constant for this reaction?

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?

A mixture of \(0.2000 \mathrm{~mol}\) of \(\mathrm{CO}_{2}, 0.1000 \mathrm{~mol}\) of \(\mathrm{H}_{2},\) and \(0.1600 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}\) is placed in a \(2.000-\mathrm{L}\) vessel. The following equilibrium is established at \(500 \mathrm{~K}\) : $$\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ (a) Calculate the initial partial pressures of \(\mathrm{CO}_{2}, \mathrm{H}_{2},\) and \(\mathrm{H}_{2} \mathrm{O} .\) (b) At equilibrium \(P_{\mathrm{H}_{2} \mathrm{O}}=3.51 \mathrm{~atm} .\) Calculate the equilibrium partial pressures of \(\mathrm{CO}_{2}, \mathrm{H}_{2},\) and \(\mathrm{CO} .\) (c) Calculate \(K_{p}\) for the reaction. (d) Calculate \(K_{c}\) for the reaction.

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