At a particular temperature, K = 1.00 times 10^2 for the reaction: In a n experi

Question

At a particular temperature, K = 1.00 times 10^2 for the reaction: In a n experiment at this temperature, 1.80 mol of H_2 and 1.80 mol of F_2 are introduced into a 1.15-L flask and allowed to react. At equilibrium, all species remain in the gas phase. What is the equilibrium concentration (in mol/L) of H_2? What is the equilibrium concentration (in mol/L) of HF? To the mixture above, an additional 6.30 times 10^-1 mol of H_2 is a added. What is the new equilibrium concentration (in mol/L) of HF?

Explanation / Answer

H2(g) + F2(g) <===> 2HF(g)

K = [HF]^2/[H2][F2]

Concentration of H2 = 1.8/1.15 = 1.565 M

concentration of F2 = 1.8/1.15 = 1.565 M

at equilibrium

concentration of HF = 2x

Concentration of H2 = 1.565-x M

concentration of F2 = 1.565-x M

1*10^2 = (2x)^2/((1.565-x)(1.565-x))

x = 1.304 M

Concentration of H2 at equilibrium = 1.565-1.304 = 0.261 M

Concentration of HF at equilibrium = 2*1.304 = 2.608 M

after addition of H2

concentration of HF = 2.608+x

Concentration of H2 = (0.261+0.63)-x M

concentration of F2 = 0.261-x M

1*10^2 = (2.608+x)^2/(((0.261+0.63)-x)(0.261-x))

x = 0.157 M

concentration of HF = 2.608+0.157 = 2.765 M

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