Question
Consider the formation of hydrogen fluoride:
H2(g) + F2(g) %u2194 2HF(g)
If a 2.0 L nickel reaction container (glass cannot be used because it reacts with HF) filled with 0.0063 M H2 is connected to a 3.5 L container filled with 0.033 M F2. The equilibrium constant, Kp, is 7.8 x 10^14 (Hint, this is a very large number, what does that imply?) Calculate the molar concentration of HF at equilibrium.
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Suppose a 4.00 L nickel reaction container filled with 0.0080 M H2 is connected to a 4.00 L container filled with 0.112 M F2. Calculate the molar concentration of H2 at equilibrium.
Explanation / Answer
Total no.of moles of H2 is , n = Molarity x Volume in L = 0.0063 M x 2.0 L = 0.0126 mol Total no.of moles of F2 is , n' = Molarity x volume in L = 0.033 M x 3.5 L = 0.1155 moles Now total volume of the reaction vessel is , V = 2.0 L + 3.5L = 5.5 L Given Kp = 7.8x10 14 We know that Kp = Kc x (RT) %u0394n Where R = gas constant = 8.314 J/mol-K T = absolute temperature = 298 K ( room temperature) %u0394n = change in no.of moles H2(g) + F2(g) <---> 2HF(g) = no.of moles of gaseous products - no.of moles of gaseous reactants = 2- (1+1) = 0 So Kc = Kp x (RT) -%u0394n = 7.8x10 14 x (8.314x298) 0 = 7.8x10 14 H2(g) + F2(g) <---> 2HF(g) initial moles 0.0126 0.1155 0 change -x -x +2x equb moles 0.0126-x 0.1155-x 2x Equilibrium constant , Kc = [HF] 2/ ( [H2][F2] ) 7.8x10 14 = (2x) 2 / [(0.0126-x ) x (0.1155 -x) ] = 4x2 / [ 0.00145 -0.1281x - x 2 ] 7.8x10 14 x [ 0.00145 -0.1281x - x 2 ] = 4x2 0.00145 -0.1281x - x 2 = 5.12x10 -15 x x 2 ( it can be neglected) So 0.00145 -0.1281x - x 2 = 0 ---> x 2 + 0.1281 - 0.00145 = 0 Solving we get x = 0.01046 mol Equilibrium moles of HF = 2x = 2x0.01046 =0.02093 mol So Equilibrium concentration of HF is ,M = No.of moles / total volume of the solution = 0.02093 mol / 5.5 L = 0.0038 M Simillarly do the second one with similar procedure