Question

Two different proteins X and Y are dissolved in aqueous solution at 37 ∘C. The proteins bind in a 1:1 ratio to form XY. A solution that is initially 1.00 mM in each protein is allowed to reach equilibrium. At equilibrium, 0.23 mM of free X and 0.23 mM of free Y remain. What is Kc for the reaction?

Answer 1

Answer:

Kc = 1.4556 * 104

Explanation:

Two different proteins X and Y will bind in a 1:1 ratio to form XY

X + Y ⇔ XY

This means that for 1 mole X consumed, there will be 1 mole of Y consumed and 1 mole of XY produced

Both proteins X and Y have initially 1mM (= 0.001M)

if there is consumed x moles of protein X then there will also be consumed x moles of protein Y. As given the remaining concentration of both proteins X and Y is 0.23 mM (= 0.23 * 10^-3 M).

This means also that here will be produced x of the protein XY

x = 10^-3 - 0.23*10-3 = 7.7 * 10^-4 M

Kc = [XY] /[X][Y]

Kc = 7.7 * 10^-4 / ((0.23 * 10^-3)*(0.23 * 10^-3))

Kc = 1.4556 * 104

Answer 2

The equilibrium constant Kc gives the equilibrium of the products and the reactant. The Kc for the reaction will be $1.45 \times 10^{4}$.

What is equilibrium constant?

Two various proteins X and Y binds in a 1:1 ratio to form XY as,

$\rm X + Y \Leftrightarrow XY$

From this, it can be said that 1 mole X and 1 mole of Y = 1 mole of XY

The initial concentration of X and Y proteins is 0.001M.

If x moles of X and Y proteins are consumed then, $0.23 \times 10 ^{ -3}$ M is the remaining concentration.

Solving for x:

$\begin{aligned}\rm x &= 10^{-3} - 0.23 \times 10^{-3} \\\\&= 7.7 \times 10^{-4} \;\rm M\end{aligned}$

Substituting values in the equilibrium constant equation:

$\begin{aligned} \rm Kc &=\rm \dfrac{[XY]}{[X][Y]}\\\\\rm Kc &= \dfrac{7.7 \times 10^{-4}}{ ((0.23 \times 10^{-3}) \times (0.23 \times 10^{-3}))}\\\\\rm Kc &= 1.4556 \times 10 ^{4} \end{aligned}$

Therefore, $1.45 \times 10^{4}$ is the equilibrium constant.

Learn more about the equilibrium constant here:

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