I keep setting this problem up but cannot get the right answer. Help would be ap
ID: 928325 • Letter: I
Question
I keep setting this problem up but cannot get the right answer. Help would be appreciated!13. At a particular temperature the equilibrium constant for the reaction:
H2(g) + F2(g) 2HF(g)
is K = 81.0. A reaction mixture in a 10.00-L flask contains 0.38 moles each of hydrogen and fluorine gases plus 0.42 moles of HF. What will be the concentration of H2 when this mixture reaches equilibrium? First you will need to determine which direction the reaction must go. Then solve as usual. The `x' you solve for will probably be the CHANGE in concentration, not the final concentration. I keep setting this problem up but cannot get the right answer. Help would be appreciated!
13. At a particular temperature the equilibrium constant for the reaction:
H2(g) + F2(g) 2HF(g)
is K = 81.0. A reaction mixture in a 10.00-L flask contains 0.38 moles each of hydrogen and fluorine gases plus 0.42 moles of HF. What will be the concentration of H2 when this mixture reaches equilibrium? First you will need to determine which direction the reaction must go. Then solve as usual. The `x' you solve for will probably be the CHANGE in concentration, not the final concentration.
Explanation / Answer
Initial concentration of F2 = H2 = number of moles / volume in L
= 0.38 moles / 10.0 L
= 0.038 M
Initial concentration of HF = number of moles / volume in L
= 0.42 mol / 10.0 L
= 0.042 M
H2(g) + F2(g) 2HF(g)
initial conc 0.038 0.038 0.042
change -a -a +2a
Equb conc 0.038-a 0.038-a 0.042+2a
Equilibrium constant , K = [HF]2 / ([H2][F2])
81.0 = (0.042+2a)2 / [(0.038-a)(0.038-a)]
[(0.042+2a) / (0.038-a) ] = 9.0
a = 0.03
There the equilibrium concentration of H2 = equilibrium concentration of F2 = 0.038 - a = 0.038-0.03 = 0.008M
The equilibrium concentration of HF = 0.042+2a = 0.042+(2x0.03) = 0.102 M