I am stuck on this chemistry problem

Question

After a 1.0 mole sample of HI (g) is placed into an evacuated 1.0 L container at 700. K, the reaction represented below occurs. The concentration of HI(g) as a function of time is shown below. 

2HI (g) ↔ H2 (g) + I2 (g)

(a) What is [HI] at equilibrium? 

(b) Determine the equilibrium concentrations of H2(g) and I2(g). 

(c) Calculate the value of the equilibrium constant at 700. K.

(d) Another chemist runs this reaction at 1,000 K, where the value of K for the reaction is 2.6 × 10-2, and measures the partial pressures of the gases. The chemist begins the experiment by filling the reaction chamber with 0.75 atm of HI(g), 0.10 atm of H2(g), and 0.50 atm of I2(g) and the reaction is allowed to reach equilibrium at 1,000 K. Determine whether the equilibrium concentration of HI(g) will be greater than, equal to, or less than the initial concentration of HI(g). Justify your answer.

Answer 1

a) We know equilibrium has occurred when the concentrations are not changing. This indicates that the rate of the forward reaction and reverse reaction are the same. Since the concentration of HI is constant 0.80 M after some time, we know that this is an equilibrium state, so [HI]_eq = 0.80 M, just from reading the graph.

b) This can be determined using stoichiometry and an ICE box. Since the concentration of HI changed from 1.00 M to 0.80M, that means I lost 0.20 M HI. This is what was used to make the products. Looking at the mole ratios in the balanced equation, we can see how for every 2 moles of HI we use, we would make one mole of H₂ and one mole of I₂. Therefore, since there is no H₂ or I₂ in the container initially, the equilibrium concentrations of H₂ and I₂ are [H₂]_eq = 0.10 M, and [I₂]_eq = 0.10 M.

c) The equilibrium constant K is determined by the ratio of products over reactants, each raised to the power of their respective coefficients. The equilibrium expression for this reaction would be:

K = [H₂][I₂] / [HI]²

... which is just pulled form the balanced equation. We plug in equilibrium concentrations into this expression to solve for K.

K = (0.10)(0.10) / (0.80²) = 0.016

d) Changing the temperature changes the value of K, so that's why the new equilibrium constant is 0.026 at this temperature.

However, this time we don't know what the equilibrium concentrations will be. We need to see IF we are at equilibrium first, and if not, WHICH WAY the reaction will shift to reach equilibrium. We are going to plug the given values into the equilibrium expression and solve for Q. Q is just a number we will use to compare to K, to see how far off we are from the expected value, don't let it confuse you.

Something that is kind of sneaky about this problem is that it gives us partial pressures in atm rather than concentrations in molarity. This means we would have to use a DIFFERENT K value. Kc is for concentration, which is what we were using above, but Kp is for partial pressures. The formula to convert between them is: Kp = Kc*(RT)^Δn, where Δn = the difference in number of moles of gas after the reaction. However, since we have 2 moles of HI gas on the reactants side, and on the products I have 1 mole of H₂ and 1 mole of I₂, this means Δn = (1 + 1) – 2 = 0, so for this reaction, Kp = Kc. Sorry, you don't actually need to worry about it this time, but it's something to watch out for in the future.

So anyway, back to plugging in. The math to solve for Q will look like this:

Q = [H₂][I₂] / [HI]² = (0.10)(0.50) / (0.75²) = 0.089

Now we compare this number to the value for K at 1,000 Kelvin. In this case, Q > K. This means that I have too many products present, so my reaction will shift left to make more reactants, so that my ratio of products over reactants equals K. This will cause more HI (the reactant) to be formed. Therefore, the concentration of HI at equilibrium will be greater than the initial concentration of HI. (There is a way to calculate exactly how much greater, but goes beyond the scope of this question.)

Hope this helps!