Equilibrium Questions

Question

After a 2.0 mole sample of HI (g) is placed into an evacuated 1.0 L container at 700. K, the following reaction occurs: 

2 HI_(g) ↔ H_2(g)  + I_2(g)

The concentration of HI_(g) as a function of time is shown below.

1. Write the expression for the equilibrium constant, K_c, for the reaction.
2. What is [HI] at equilibrium?
3. Determine the equilibrium concentrations of H_2(g) and I_2(g).
4. On the graph above, make a sketch that shows how the concentration of H_2(g) changes as a function of time.
5. Calculate the value of the following equilibrium constants for the reaction at 700.K
6. Kc
7. _Kp
8. At 1,000 K, the value of Kc for the reaction is 2.6 x 10-2. In an experiment, 0.75 mole of HI(g), 0.50 mole of H2(g), and 0.30 mol of I2(g) are placed in a 1.0 L container and 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

1. Kc = [H₂][I₂]/[HI]²
2. [HI] at equilibrium = 1.60 M
3. Since equilibrium [HI] is 1.60, [H₂] & [I₂] = 0.20 M
4. Unable to sketch graph on this platform
5. Kc = (0.2)(0.2)/(1.6)² = 0.0156 and for Kp you have .......... Kp = KC(RT)^Δn = 0.0156 x 57.5 = 4.7
6. Q = (0.5)(0.3)/(0.75)² = 0.27. Under these conditions, Q > Kc, therefore reaction will run to the left toward the reactant, so equilibrium [HI] will be greater than initial concentration.