Question

1.The equilibrium constant, Kc, for the reaction A(g) + B(g)  AB(g) is 115 at 25 oC. If 1.580 x 10‐1 mol of A and 4.721 x 10‐5 mol of B are introduced into a 1.00 liter vessel at 25 oC, calculate the equilibrium concentration of all species. Show your work using an ICE table.

2. (a) This experiment is separated into parts I and II. What are the [Fe3+] and [SCN‐] before mixing in each part?

(b) Briefly, explain why the concentrations of the Fe3+ and SCN‐ are different in the two parts of the experiment.

Answer 1

To solve the problem 1, we have to write the balanced reaction and solve a ICE table. The balanced reaction is given. Initially we have 1.580 x 10‐1 mol of A and 4.721 x 10‐5 mol of B in a 1 liter vessel at 25°C. We can solve that problem only at 25°C because the value of Kc is at 25°C. To built the ICE chat we have to use all concentration in molarity, how the vessel has a capacity of 1 liter, we can use directly the moles of both reagents given. In the second step (change) we have to substract the moles of each reagents thtas reacts, in our case is X, because, in the balanced reaction, we have 1 mol of A and 1 mol of B, and appears X moles of AB, because in the balanced reaction we have 1 mol of AB. So, stoichiometric coefficients says us the amount of X for each compound.

A(g) + B(g) ---------- AB(g)

I) 1.580 x 10‐1 moles   4.721 x 10‐5 mol 0

C) -X -X X

E) 1.580 x 10‐1 - X 4.721 x 10‐5 - X 0 + X

now we have to use the concentration in equilibrium in the Kc expression, Kc expression is formed by the concentration of the products (AB in our case) elevated to its stoichiometric coefficients, (1 in our case) over the concentration of the reagents (A and B in our case) elevated to its stoichiometric coefficients (1 and 1 in our case)

so,

now we have to put our concentration oa all species in equilibrium

and now we have to solve to find X. (the lowercase x indicates multiplication)

applying distributive property, we have:

Kc (0.158 - X) x (4.721 x 10‐5 - X) = X

(0.158 KC - XKc).(4.721 x 10‐5 - X) = X

applying distributive property again, we have:

4.721 x 10‐5 x 0.158 x Kc - 0.158 x X x Kc - 4.721 x 10‐5 x X x Kc + X^2 x Kc = X

8.578x10^-4 - 18.17X - 5.43x10^-3X + 115X^2 - X = 0

115X^2 - 19.175X + 8.578x10^-4 = 0

now, solving that quadratic equation we find 2 values of X, (quadratic equation has 2 solutions)

X1=0.167

X2=4.47x10^-5

X1 is not correct, because we use it, the concentrations of B in equilibrium would be negative, and that is impossible.

X2 is the correct value, so, using it we can calculate the concentration of all species in equilibrium.

for A:

[A] = 1.580 x 10‐1 - X

   = 1.580 x 10‐1 - 4.47x10^-5

[A] = 0.15795 M

for B:

    [B] =  4.721 x 10‐5 - X   

= 4.721 x 10‐5 - 4.47x10^-5

  [B] = 2.46x10^-6 M

for AB

[AB] = X

[AB] = 4.4747x10^-5 M

Problem 2 isnt clear if it ie referred to problem 1 or not, really i dont understand. Sorry