hi

At a particular temperature, K = 1.00 102 for the following reaction.
H2(g) + I2(g)= 2 HI(g)

In an experiment, 1.73 mol H2, 1.73 mol I2, and 1.73 mol HI are introduced into a 1.00 L container. Calculate the concentrations of all species when equilibrium is reached.
H2

okay so i set up an ICE table

and the HI at equillibrium would be

1.73+2x M right?? i got the concentration of the reactants i just need help with the HI especially when there is two as the coefficient!

thanks!

It seems to me that if K = 1.00 that the final concentrations will be 1.73 M for each since

Kc = 1.00 = (HI)^2/((H2)(I) = (1.73)^2/(1.73)(1.73) = 1.00

That's correct. Just play around with the number of sig figs (there should be 3) and your answer will be correct.

At what temperature will be the total kinetic energy of 0.30 moles of He be the same as total kinetic energy of 0.40 mole of Ar at 400 K ?

Hi! I can help you with that.

First, let's set up the ICE table for the reaction:

H2(g) + I2(g) → 2 HI(g)

H2(g) + I2(g) → 2 HI(g)
Initial: 1.73 mol 1.73 mol 1.73 mol
Change: -x -x +2x
Equilibrium: 1.73 - x 1.73 - x 1.73 + 2x

Now, let's focus on the concentration of HI at equilibrium. As you correctly mentioned, the coefficient of HI in the balanced equation is 2. This means that for every 1 mole of HI that reacts, 2 moles of HI are formed.

So, at equilibrium, the expression for the concentration of HI can be written as:

[HI] = 1.73 + 2x

Since [HI] represents the concentration of HI at equilibrium, set it equal to K and solve for x:

K = [HI]^2 / ([H2][I2])

1.00 x 10^2 = (1.73 + 2x)^2 / ([H2][I2])

Now, plug in the initial concentrations of H2 and I2 (both are 1.73 mol) into the denominator:

1.00 x 10^2 = (1.73 + 2x)^2 / (1.73)(1.73)

Simplify the equation:

1.00 x 10^2 = (1.73 + 2x)^2 / 2.9929

Cross-multiply:

1.00 x 10^2 * 2.9929 = (1.73 + 2x)^2

Then, take the square root of both sides:

√(1.00 x 10^2 * 2.9929) = 1.73 + 2x

Solve for x:

x = (√(1.00 x 10^2 * 2.9929) - 1.73) / 2

Once you calculate the value of x, substitute it back into the expression for [HI] to find the concentration at equilibrium:

[HI] = 1.73 + 2x

Now you have the concentration of HI at equilibrium.