I need help on part b please! (and if you want to double check me on HF of part

Question

I need help on part b please! (and if you want to double check me on HF of part A)

At a certain temperature, K = 1.00 102 for the reaction given below. H2(g) + F2(g) 2 HF(g) (a) Calculate the concentrations of all species in the equilibrium mixture produced by adding 3.20 mol of H2 and 3.20 mol of F2 to a 1.00 L container.

H2 .53 M (correct)

F2 .53 M (correct)

HF 1.1 M (incorrect: wasn't thinking and multiplied .53x2 rather than 2.67x2. I believe this should have been 5.34)

(b) What would the equilibrium concentration of HF be if 1.30 mol of HF is removed from the equilibrium mixture in Part a?

Explanation / Answer

H2(g) + F2(g) -----> 2HF(g) (g)

3.2-x 3.2-x 2x

Ka = (2x)^2/(3.2-x)(3.2-x) = 100

so x = 2.666 M

so

[H2] = 3.2-2.67 = 0.53 = [F2]

[HF] = 2x = 2*2.67 = 5.34 M

part 2

H2(g) + F2(g) -----> 2HF(g) (g)

3.2-x 3.2-x 2x-1.3

Ka = (2x-1.3)^2/(3.2-x)^2 = 100

x = 2.775

[H2] = [F2] = 3.2-2.775 = 0.425 M

[HF] = 2x-1.3 = 4.25 M

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