Problem 39 - Chemistry The Central Science In Si Units

Question

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 XY. A solution that is initially \(1.00 \mathrm{mM}\) 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 Y remain. What is \(K_{c}\) for the reaction?

Step-by-Step Solution

Answer

The equilibrium constant \(K_c\) is 20.

Step 1: Understand the Reaction

We begin with the reaction equation: X + Y ↔ XY. Initially, there is 1.00 mM of each protein, X and Y.

Step 2: Set Up ICE Table

Write the Initial, Change, and Equilibrium concentrations for X, Y, and XY.\[\begin{array}{c|ccc}\text{Species} & \text{Initial (mM)} & \text{Change (mM)} & \text{Equilibrium (mM)} \\hline\text{X} & 1.00 & -x & 0.20 \\text{Y} & 1.00 & -x & 0.20 \\text{XY} & 0 & +x & x \\end{array}\]Since 0.20 mM of X and Y remain, x = 1.00 - 0.20 = 0.80 mM is the concentration of XY at equilibrium.

Step 3: Write the Expression for Kc

The equilibrium constant expression for the reaction is:\[K_c = \frac{[XY]}{[X][Y]}\]We substitute the equilibrium concentrations into this expression.

Step 4: Calculate the Kc Value

Substitute the equilibrium concentrations into the Kc expression:\[K_c = \frac{0.80}{0.20 \times 0.20}\]Calculate \(0.20 \times 0.20 = 0.04\), so:\[K_c = \frac{0.80}{0.04} = 20\]

Step 5: Conclude the Answer

Thus, the equilibrium constant \(K_c\) for the reaction is 20.

Key Concepts

Equilibrium Constant

The equilibrium constant, often represented by the symbol \( K_c \), is a crucial concept in understanding reactions at equilibrium. It quantifies the ratio of concentrations of products to reactants at equilibrium. For this specific protein binding reaction, where protein X binds with protein Y to form a complex XY, the equilibrium constant expression is given by:

• \( K_c = \frac{[XY]}{[X][Y]} \)

Notice how the product concentration \([XY]\) is in the numerator, while the reactants \([X]\) and \([Y]\) are in the denominator. This ratio remains constant for a given reaction at a specific temperature. In our exercise, the calculated \( K_c \) is 20, indicating that at equilibrium, the formed product XY is favored.

Protein Binding

Protein binding is a fascinating biological process where two or more protein molecules interact to form a complex. This is essential for many biological functions, including enzyme actions, signaling, and structural activities. In our exercise, proteins X and Y interact in a 1:1 ratio to form a new compound, XY.

• Proteins X and Y initially have equal concentrations of 1.00 mM.
• They combine to form a stable complex, altering the concentrations of the free proteins.
• At equilibrium, only 0.20 mM of each protein remains unbound, showing how efficiently they bind to each other.

The strength and likelihood of protein binding can be influenced by several factors, including temperature, presence of other molecules, and specific 3D structures of proteins.

ICE Table

An ICE table is a structured way to organize data concerning chemical reactions. ICE stands for Initial, Change, and Equilibrium.

• **Initial:** This column records the starting concentrations of reactants and products.
• **Change:** Here, we account for the increase or decrease in concentrations as the reaction progresses towards equilibrium.
• **Equilibrium:** This column shows the concentrations when the system has reached equilibrium.

For our reaction, initially, both proteins had a concentration of 1.00 mM. After computing the changes, the equilibrium concentrations revealed 0.80 mM of product XY, and 0.20 mM each of the unbound proteins X and Y. The ICE table is an invaluable tool for visualizing and calculating the shifts in concentrations during a reaction.

Equilibrium Concentrations

Equilibrium concentrations refer to the levels of reactants and products when a chemical reaction has reached equilibrium. At this state, the rate of the forward reaction equals the rate of the reverse reaction, so there is no net change in concentrations.

• In our protein binding exercise, the equilibrium concentrations of free proteins \( X \) and \( Y \) are both 0.20 mM.
• The concentration of the protein complex \( XY \) is 0.80 mM at equilibrium.

Knowing these concentrations allows us to calculate the equilibrium constant. Understanding these values helps predict how changes in conditions might affect the overall equilibrium and hence, the amounts of each substance present in the reaction mixture.