Fred performed the following equilibrium experiment: In a 2.50 L flask at (530 d

Question

Fred performed the following equilibrium experiment: In a 2.50 L flask at (530 degree C) the following chemical species were present: 49.98 grams of HF gas, 0.50 grams of H2 gas and 14.24 grams of Fz gas. Fred knows that the K_c for this reaction at this temperature is 1.00X10^-2. The balanced chemical equation for this reaction is: 2HF (g) r H_2 (g) + F_2(g) Given the above experimental conditions and information, help Fred answer the following questions: Is this reaction is at equilibrium? YES or NO You must show work to justify answer. If it is not at equilibrium, then predict the direction the reaction would shift to obtain equilibrium. Determine the equilibrium concentrations for each of the reaction components based on your prediction in What is the KP value for this reaction? this reaction product or reactant favored? State your reasoning for your response.

Explanation / Answer

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

Keq = {[H2]*[F2]}/[HF]2

Now, molar mass of H2 = 2 g/mole

molar mass of F2 = 38 g/mole

molar mass of HF = 20 g/mole

Now, moles of HF in 49.98 g of it = mass/molar mass = 2.5

moles of H2 in 0.5 g of it = mass/molar mass = 0.5/2 = 0.25

moles of F2 in 14.24 g of it = mass/molar mass = 14.24/38 = 0.375

Thus, reaction quotient,Kq = (0.375*0.25)/2.52 = 0.015

Since Kq > Kc , therefore the reaction is not in equilibrium

The reaction moves in the backward direction, i.e formation of reactants is favoured

Kp = Kc*(R*T)n ; where n = moles of gaseous products - moles of gaseous reactants = 0

Thus, Kp = KC = 1*10-2

Since Kc value <1 , the reaction is reactant favoured.

Let at eqb., moles of HF = 2.5+2x and F2 = (0.375-x), H2 = (0.25-x)

Thus, 0.01 = (0.25-x)*(0.375-x)/(2.5+2x)2

or, x = 0.07

Thus, equilibrium [HF] = 2.64/2.5 = 1.056 M

[H2] = 0.18/2.5 = 0.072 M

[F2] = 0.305/2.5 = 0.122 M

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