3A + 2B ==> C + D
Set up an ICE chart.
Initial:
A = 0.0392 M
B = 0.0452 M
C = 0 M
D = 0 M
part of equilibrium:
C = 0.00128 M
change:
Therefore, C must have changed by + 0.00128.
D must have changed by + 0.00128
A must have changed by -3*0.00128
B must have changed by -2*0.00128
remainder of equilibrium:
A is now 0.0392-change in A.
B is now 0.0452-change in B.
C is 0.00128
D is 0.00128
Now set up Keq expression, plug in the value and solve for Kc.
A mixture containing 0.0392 M A(g) and 0.0452 M B9g) is allowed to come to equilibrium at 300K.
The reaction: 3A(g)+ 2B(g)=> C(g)+ D(g) occurs. At equilibrium, [C]=0.00128 M
What is the value of Kc?
1 answer
1 answer