8 pts 23 The equilibrium constant K for the reaction below i

(8 pts) 23. The equilibrium constant K, for the reaction below is 4.31 x 104 at 375 C. F2(g) + H2(g) 2 HF(g) In a certain experiment a student starts with 0.862 atm of F2 and 0.373 atm of H2 in a constant volume vessel at 375°C. Calculate the partial pressures of all species when equilibrium is reached. (6 pts) 24. Write equilibrium expressions for the following reactions. Label each reaction as a homogeneous equilibrium or a heterogeneous equilibrium. 4 HCI(g) + O2(g) 2 Cla(g) + 2H20g) a. b. LaCl3(6) + H20(g) (10 pts)25. For the reaction below, state the direction (right, left, or no change) in which the equilibrium will shift if the following conditions are applied. a. add chlorine b. remove oxygen c. decrease the pressure d. increase the temperature e. add a catalyst

Solution

23.
     F2(g) + H2(g) <----------> 2HF(g)
I   0.862    0.373               0
C    -x       -x                 2x
E   0.862-x 0.373-x             2x

    Kp = P^2HF/PF2PH2
    4.31*10^-4 = (2x)^2/(0.862-x)(0.73-x)
   4.31*10^-4*(0.862-x)(0.73-x) = 4x^2
    x = 0.0081
   PF2 = 0.862-x = 0.862-0.0081   = 0.854atm
   PH2 = 0.373-x = 0.373-0.0081    = 0.365atm
   PHF = 2x       = 2*0.0081        = 0.0162atm

24. In the equilibrium reactant and products are same phage is called homogeneous equilibrium
    4HCl(g) + O2(g) <----------> 2Cl2(g) + 2H2O(g) is homogeneous equilibrium
In the equilibrium reactant and products are different phage is called heterogeneous equilibrium
      b. LaCl3(s) + H2O(g) ---------> LaOCl(s) + 2HCl(g) is heterogeneous equilibrium
25.
    4HCl(g) + O2(g) --------> 2Cl2(g) + 2h2O(g)   DH = -114KJ

a. add chlorine - left
b. remove oxygen - left
c. decrease the pressure - left
d. increases the temperature - left
e. add a catalyst - no change