initial concn = 0.067/0.250 = approx 0.27 but that's just an estimate.
...........A ==> B
I.......0.027...0
C.........-x....x
E......0.027-x..x
Substitute the E line into Kc expression and solve for x.
The value of KC for the interconversion of butane and isobutane is 2.5 at 25°C. If you place 0.067 mol of butane in a 0.250-L flask at 25°C and allow equilibrium to be established, what will be the concentrations of the two forms of butane?
A)[butane]
B)[isobutane]
1 answer