What is the equilibrium concentration of I2 in the gaseous mixture? H2(g) + I2(g) ↔ 2HI?

Question

H2(g) + I2(g) ↔ 2HI Temperature = 731K




2.10 mol of H2 and 2.10 mol of I2 are placed in a 1.00 L vessel. What is the equilibrium concentration of I2 in the gaseous mixture?
The equilibrium constant is K = 49.0


***HINT Normally you would need to solve a quadratic equation here since K is large.
However, in this case the intial concentrations of the two reactants are the same.
You can set up the equilbirium ratio (products over reactants) in terms of x, set it equal to K in the usual way,
then take the square root of both sides.
Then solve for x in the resulting linear equation.
OR you can solve the quadratic equation, if you prefer.

Answer 1

Your question gives you all of the ingredients to solve the problem.

K(eq) = (HI)E2 / (H2) (I2), where the quantitites in brackets is the concentration. SInce the reactants have the same concentration, the equation reduces to K(eq) = (HI)E2 / (H2)E2

You have the K(eq) and you have the (H2). Just plug in, solve for (HI)E2, take the square root and solve.

Answer 2

permit X = moles of H2 and I2 that disappear to style hi throughout the reaction. The equilibrium concentrations of H2 and I2 at equilibrium would be (2-X)/a million.06 and (3.6-X)/a million.06, and the concentration of hi at equilibrium would be 2X/a million.06. That make experience so far? Now, basically plug all that into your expression for ok and remedy for X. you will see that the a million.06's will all basically cancel out, and you would be wanting: 40 9.7 = (2X)^2/(2-X)(3.6-X) It should not be exciting to remedy, and you will would desire to apply the quadratic equation, yet you will desire to be waiting to do it...