Initially, 0.980 mol of A is present in a 2.00 L solution.

Question

Initially, 0.980 mol of A is present in a 2.00 L solution.

2 A(aq) ⇌ 2 B(aq) + C(aq)

At equilibrium, 0.110 mol of C is present. Calculate K.

K=

Answer 1

2A <==> 2B + C

0.980......0......0.....Initial

-2x........+2x...+x....Change

0.98-2x...2x....x.....Equilibrium

Since 0.110 mol C present @ equilibrium, that means x = 0.110

So, at equilibrium we have the following moles of each:

A = 0.98 - 2x0.110 = 0.760 mols

B = 2x = 2x0.110 = 0.220 mols

C = 0.110 mols

Since the volume is 2.00 L, we have the following concentrations @ equilibrium:

[A] = 0.760 mol / 2.00 L = 0.380 M

[B ]= 0.220 mol / 2.00 L = 0.110 M

[C] = 0.110 mol / 2.00 L = 0.0550 M

K = [B]²[C] / [A]²

K = (0.110)²(0.0550) / (0.380)²

K = 0.00461