(HCHO) = (H2) = (CO) = 1 mol/2L = 0.5 M
add (HCHO) = 2/2 = 1.0M
add (CO) = 1/2 = 0.5M
..........HCHO(g) --> H2(g) + CO(g)
E........0.5M........0.5M.....0.5M
add.......1.0M..........0......0.5M
Substitute the E line into Kc expression and calculate Kc.
#2.
a. Since the system is not at equilibrium #1 isn't true.
b. Right, the rxn mixture is not at equilibrium BUT further reaction WILL occur.
c. No, the rxn mixture will MOVE to equilibrium
d & e.One of these is true and the other false. Which way will the reaction shift to reattain equilibrium? Solve for the reaction quotient and compare it with Kc you obtained from #1.
Consider the reaction
HCHO(g) *) H2(g) + CO(g).
1.0 mol of HCHO, 1.0 mol of H2 and 1.0 mol
of CO exist in equilibrium in a 2.0 L reaction
vessel at 600�C.
a) Determine the value of the equilibrium
constant Kc for this system.
2.0 moles of HCHO and 1.0 mol of CO are
then added to this system.
b)Which of the following statements about
this reaction is now true?
1. The forward rate of this reaction is the same as the reverse rate at these new concentrations.
2.The reaction mixture is not at equilibrium, but no further reaction will occur.
3. The reaction mixture remains at equilibrium.
4. The reaction mixture is not at equilibrium, but will move toward equilibrium by using up more HCHO.
5. The reaction mixture is not at equilibrium, but will move toward equilibrium by forming more HCHO.
1 answer
1 answer