Knowee
Questions
Features
Study Tools

A solution is made by mixing 30.0 mL of 0.150M (mol/L) compound A with 25.0 mL of 0.200M (mol/L) compound B. At equilibrium, the concentration of C is 0.0454M. Calculate the equilibrium constant, K, for this reaction.A (aq) + 2B (aq) <---> C (aq)The initial concentrations of A and B are not 0.150 M and 0.200 M

Question

A solution is made by mixing 30.0 mL of 0.150M (mol/L) compound A with 25.0 mL of 0.200M (mol/L) compound B. At equilibrium, the concentration of C is 0.0454M. Calculate the equilibrium constant, K, for this reaction.A (aq) + 2B (aq) <---> C (aq)The initial concentrations of A and B are not 0.150 M and 0.200 M

🧐 Not the exact question you are looking for?Go ask a question

Solution

To calculate the equilibrium constant, K, for this reaction, we need to use the concentrations of the reactants and products at equilibrium.

Given:

  • Volume of compound A (A) = 30.0 mL
  • Concentration of compound A (A) = 0.150 M
  • Volume of compound B (B) = 25.0 mL
  • Concentration of compound B (B) = 0.200 M
  • Concentration of compound C (C) at equilibrium = 0.0454 M

The balanced equation for the reaction is: A (aq) + 2B (aq) ⇌ C (aq)

To calculate the equilibrium constant, K, we need to use the formula: K = [C]^c / ([A]^a * [B]^b)

Where [C], [A], and [B] represent the concentrations of C, A, and B at equilibrium, respectively, and a, b, and c are the stoichiometric coefficients of the balanced equation.

In this case, the stoichiometric coefficients are: a = 1 (for A) b = 2 (for B) c = 1 (for C)

Plugging in the values, we have: K = (0.0454) / ((0.150)^1 * (0.200)^2)

Calculating this expression, we find the value of K.

This problem has been solved

Similar Questions

Samples of A (1.0 mol) and B (3.5 mol) are placed in a 1 L container and the following reaction takes place:A(g)  +  B(g)  ⇄ 2C(g)  At equilibrium, the concentration of A is 0.5 M.  What is the value of K to 2 d.p?

1.1 mol of A mixed with 2.2 mol of B and the mixture is kept in a 1 L flask and the equilibrium,A + 2B  2C + D is reached. If at equilibrium 0.2 mol of C is formed then the value of KC will be.

For the reaction    A(g)  +  B(g)  ⇋   2 C(g) the following equilibrium concentrations were found:[A(g)]  =  0.358 mol / L[B(g)]  =  0.216 mol / L[C(g)]  =  0.750 mol / LCalculate the value of the equilibrium constant Kc. Give 3 significant figures in your answer.

For the reaction, A + B⇌C + D, the initial concentration of ‘A’ and ‘B’ are equal, but the equilibrium concentration of ‘C’ is twice that of equilibrium concentration of ‘A’. Theequilibrium constant is _______

For the reaction:2 A (g) B (g) + C (g)at 900 ° C, Kc is 1.40 x 10 -3. If 0.780 mole of A (g) and 0.244 mole each of B (g) and C (g) are mixed in a1.00 liter container at 900 ° C, what are the concentrations of A, B, and C at equilibrium?

1/3

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.