Question

Given the equation: 3A(aq) + 2B(aq) = 3C(aq) + 2D(aq) When 45.0 mL of 0.050M do A in mixed with 25.0 mL of 0.10M of B, the concentration, at equilibrium, of C was found to be 0.021M and D at 0.014M. Calculate Kc. {Tip: Moles of A that reacted = moles of C thats produced and moles of B that reacted = moles of D produced at equilibrium. From their molarities, use the total volume of the mixture. number of moles(n)= M x V in liters.} Please pay attention to what the initial concentrations and the final concentrations are when calculating Kc

Answer 1

Initial concentration of A = 45×0.050 = 2.25 mole
Initial concentration of B = 25×0.10 = 2.5 mole
Equilibrium concentration of A = 2.25 - 0.021 = 2.229 mole
Equilibrium concentration of B = 2.5 - 0.014 = 2.486 mole
Equilibrium constant Kc = (0.021)^3 × (0.014)^2 / (2.229)^3 × (2.486)^2
= 1.81 × 10^-9 / 68.44
= 0.026 × 10^-9
= 2.6 × 10^-11