Question

At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3.

H2(g)+I2(g) <--> 2HI(g)

Kc= 53.3

At this temperature, 0.300 mol of H2 and 0.300 mol of I2 were placed in a 1.00-L container to react. What concentration of HI is present at equilibrium?

HI = ? M

Please post an answer too if possible.

Answer 1

0.300mol/1.00L = 0.300M
..........H2 + I2 ==> 2HI
I......0.300..0.300....0
C.........-x....-x.....+2x
E.....0.300-x..0.300-x..2x

Substitute the E line into Kc expression and solve for x. Remember HI is 2x.