Question

What is the equilibrium constant for the reversible reaction in aqueous medium below given that respective concentrations of A, B, C, and D are 0.0117 M, 0.00440 M, 0.00550 M, and 0.00780 M? 3A + 3B 2C + 3D Report your answer to the nearest whole number.

Answer 1

The equilibrium constant for the reaction is 1315.

Step-by-step explanation:

The equilibrium constant for a reversible reaction is a measure of how far a reaction proceeds before reaching equilibrium. In this case, the reaction is 3A + 3B → 2C + 3D. The equilibrium constant, K, is calculated using the concentrations of the products divided by the concentrations of the reactants, each raised to the power of their respective stoichiometric coefficients.

K = [C]²[D]³ / [A]³[B]³

Substituting the given concentrations into the equation, we get: K = (0.00550)²(0.00780)³ / (0.0117)³(0.00440)³. Evaluating this expression gives us K = 1315.

Answer 2

The equilibrium constant for this reaction is approximately 0.41 (rounded to nearest whole number).

How to find equilibrium constant?

To calculate the equilibrium constant (Kc) for this reaction, we can use the mass action law:

`Kc = (([C]^2 * [D]^3))/([A]^3 * [B]^3)`

where:

Kc = equilibrium constant

[C] = concentration of C

[D] = concentration of D

[A] = concentration of A

[B] = concentration of B

Plug in the given concentrations:

[C] = 0.00550 M

[D] = 0.00780 M

[A] = 0.0117 M

[B] = 0.00440 M

Calculate Kc:

`Kc = (([0.00550]^2 * [0.00780]^3))/([0.0117]^3 * [0.00440]^3))`

Kc ≈ 0.408

Therefore, the equilibrium constant for this reaction is approximately 0.41 (rounded to nearest whole number).