Is the System at Equilibrium with Given Concentrations?

[brake4country]

Homework Statement

Consider the following reaction:
H₂ + I₂ → 2HI (all gases)
At 740 K, this reaction has an equilibrium constant of 50. If, at 740 K, the concentration of hydrogen gas is 2 M, iodine gas is 3 M, and hydrogen iodide gas is 6 M:
(A) the concentration of hydrogen iodide will increase over time
(B) the concentration of hydrogen iodide will decrease over time
(C) the system is at equilibrium
(D) this combination of concentrations is inconsistent with the information given

Homework Equations

NA

The Attempt at a Solution

I believe the correct answer is (A) but I wanted to make sure my reasoning is correct.
Since all are gases, the K_eq = [HI]²/[H] = 6²/2x3, therefore, K = 6. If the calculated value for K is less than the equilibrium constant of 50 in the equation, this means that the reaction will shift to produce more HI. Additionally, if the calculated K value was greater than 50, then I am assuming the the products would shift toward the left to yield more reactants.

 

[Quantum Defect]

Homework Statement

Consider the following reaction:
H₂ + I₂ → 2HI (all gases)
At 740 K, this reaction has an equilibrium constant of 50. If, at 740 K, the concentration of hydrogen gas is 2 M, iodine gas is 3 M, and hydrogen iodide gas is 6 M:
(A) the concentration of hydrogen iodide will increase over time
(B) the concentration of hydrogen iodide will decrease over time
(C) the system is at equilibrium
(D) this combination of concentrations is inconsistent with the information given

Homework Equations

NA

The Attempt at a Solution

I believe the correct answer is (A) but I wanted to make sure my reasoning is correct.
Since all are gases, the K_eq = [HI]²/[H] = 6²/2x3, therefore, K = 6. If the calculated value for K is less than the equilibrium constant of 50 in the equation, this means that the reaction will shift to produce more HI. Additionally, if the calculated K value was greater than 50, then I am assuming the the products would shift toward the left to yield more reactants.

This is correct, but usually, the non-equilibrium equivalent of Keq is called Q. Since Q (=6) < Keq, products will form, reactants will decrease, until Q = Keq -- i.e. equilibrium has been reached.

 

[brake4country]

Ok. So I basically calculated a Q value. Thank you!

 

[Borek]

For the record: Q is called a "reaction quotient".