Question

For the following reaction: H2(g) + I2(g)<--->2HI(g)     ; Kc=49.5

The beginning concentrations are: H2: 0.10M, I2: 0.050M, and HI:0

Assume this reaction is not at equilibrium. When it does reach equilibrium, what will the concentrations be?

Answer 1

Let's write again the overall reaction, and then, do a ICE chart.

r: H_2(g) + I_2(g)<-------->2HI_(g)

i: 0.2 0.05 0

e. 0.2-x 0.05-x 2x

49.5 = (2x)² / (0.2-x)(0.05-x)

From here, we solve for "x" to calculate the concentration of the reactants and product in the equilibrium:

49.5 = 4x² / 0.01 - 0.25x + x²

49.5(0.01 - 0.25x + x²) = 4x²

0.495 - 12.375x + 49.5x² = 4x²

45.5x² - 12.375x + 0.495 = 0

From here, we use the quadratic equation formula to get the two possible values of x. The lowest and positive value would be the value we'll take:

x = 12.375 (12.375² - 4*45.5*0.495)^1/2 / 2*45.5

x = 12.375 7.94 / 91

x₁ = 0.049 M

x₂ = 0.223 M

we will take x₁ = 0.049 M

Now we can calculate the concentration in equilibrium of each compound:

[H₂] = 0.2 - 0.049 = 0.151 M

[I₂] = 0.05 - 0.049 = 0.001 M

[HI] = 2*0.049 = 0.098 M

Finally, the reaction will reach it's equilibrium when the value of Kc = 0.

Hope this helps