Question

Consider the equation A(aq) + 2B(aq) 3C(aq) + 2D(aq). In one experiment, 45.0 mL of 0.050 M A is mixed with 25.0 mL 0.100 M B. At equilibrium the concentration of C is 0.0410 M. Calculate K. please be detailed towards the end on how you arranged you equation.

Answer 1

K =([C]^3*[D]^2)/([A]*[B]^2)

Total volume = 70ml

C = 0.0410M = 0.00287 moles
C: D = 2: 3
D = 0.001913 moles = 0.0273 M.

Initially A = 0.00225 moles.
A: C = 1: 3
Converted 0.000957 moles of A, leaving 0.001293 moles of A, now in 70 ml
Concentration A = 0.0185M

Initially B = 0.0025 moles
B: C = 2: 3
Converted 0.001913 moles of B, leaving 0.000587 moles of B, now in 70 ml
Concentration B = 0.0084

Then plug into top equation:

K =([0.0410]^3 *[0.0273]^2) / ([0.0185] *[0.0084]^2 )

This gives a constant of 0.040