Consider the reaction A+B->C+3D. A solution was prepared by mixing 50 ml of .001 M of A, 100 ml of .002 M of B, 10 ml of 1 M of C, and 75 ml of .0015 M of D. At equilibrium, the concentration of D was measured and found to be .0006 M. Calculate the equilibrium concentrations of A, B, C, and D. Calculate the equilibrium constant for the reaction.
total volume = 100 + 50 + 10 + 75 = 235 mL.
In the final solution at mixing and before equilibrium
(A) = 0.001M x (50/235) = ?M
(B) = 0.002M x (100/235) = ?M
(C) = 1M x (10/235) = ?M
(D) = 0.0015M x (75/235) = ?M
So you know at equilibrium (D) = 0.0006M.
Therefore, the reaction formed 1/3 * 0.0006M for C. Add that to (C) above.
(A) at equilibrium = initial concn - 1/3 * 0.0006 and (B) at equilibrium = initial concn B - 1/3 * 0.0006.
Plug those values into Keq expression and solve for Keq.
To calculate the equilibrium concentrations of A, B, C, and D, we need to use the stoichiometry of the reaction as well as the initial concentrations and volumes of the solutions.
Let's start by calculating the moles of A, B, C, and D based on their concentrations and volumes:
Moles of A = Concentration of A * Volume of A solution
= 0.001 M * 50 ml
= 0.05 mmol
Moles of B = Concentration of B * Volume of B solution
= 0.002 M * 100 ml
= 0.2 mmol
Moles of C = Concentration of C * Volume of C solution
= 1 M * 10 ml
= 10 mmol
Moles of D = Concentration of D * Volume of D solution
= 0.0015 M * 75 ml
= 0.1125 mmol
Now, let's consider the stoichiometry of the reaction A + B -> C + 3D. We can use the balanced equation to determine the change in moles of A, B, C, and D at equilibrium:
From the balanced equation, we see that for every 1 mole of A reacted, 1 mole of C is produced, and for every 1 mole of B reacted, 3 moles of D are produced.
So, at equilibrium:
Change in moles of A = -1 * Change in moles of C
Change in moles of B = -1/3 * Change in moles of D
Let's denote the change in moles of C as x,
so the change in moles of A is also -x.
Since 1 mole of C is produced for every mole of A that reacts, the change in moles of C is also equal to the change in moles of A.
Therefore:
Change in moles of A = -x
Change in moles of B = -1/3 * 3x = -x
Change in moles of C = x
Change in moles of D = 3x
Now, we can calculate the equilibrium concentrations of A, B, C, and D.
Equilibrium concentration of A = (Initial moles of A + Change in moles of A) / Total volume
= (0.05 mmol + (-x mmol)) / (50 ml + 100 ml)
= (0.05 - x) / 150 mmol/ml
Equilibrium concentration of B = (Initial moles of B + Change in moles of B) / Total volume
= (0.2 mmol + (-x mmol)) / (50 ml + 100 ml)
= (0.2 - x) / 150 mmol/ml
Equilibrium concentration of C = (Initial moles of C + Change in moles of C) / Total volume
= (10 mmol + x mmol) / (50 ml + 100 ml + 10 ml)
= (10 + x) / 160 mmol/ml
Equilibrium concentration of D = (Initial moles of D + Change in moles of D) / Total volume
= (0.1125 mmol + 3x mmol) / (50 ml + 100 ml + 75 ml)
= (0.1125 + 3x) / 225 mmol/ml
Next, let's calculate the equilibrium constant for the reaction using the equilibrium concentrations of A, B, C, and D:
Equilibrium constant (K) = ([C]^1 * [D]^3) / ([A]^1 * [B]^1)
= ([C] * [D]^3) / ([A] * [B])
Substituting the equilibrium concentrations we calculated earlier, we get:
K = ((10 + x) / 160) * ((0.1125 + 3x) / 225) / ((0.05 - x) / 150) * ((0.2 - x) / 150)
Simplifying this expression will give you the equilibrium constant for the reaction.