A 100 mL container is filled with 0.01 mol of A2B3. At equilibrium, it also contains 0.006 mol of B. Find the value of K.
A2B3(g) --> 2A(g) +3B(g)
........A2B3 ==> 2A + 3B
I......0.01.......0....0
C.......-x........2x...3x
E....0.01-x.......2x...3x
The problem tells us that 3x = 0.006 = mols B; therefore, x = 0.006/3 = 0.002
Equilibrium mols are as follows:
A2B3 = 0.01-0.002 = ?
A = 2*0.002 = ?
B = 0.006
These are mols. You must convert to molarity. Do that by M = mols/L. The volume is given as 0.1L. Then substitute concns into the Keq expression and solve for Kc.
To find the value of K, we need to use the equation for the equilibrium constant.
K = ([A]^2 * [B]^3) / [A2B3]
First, let's calculate the concentrations of A and B at equilibrium.
Given:
Initial moles of A2B3 = 0.01 mol
Initial volume of the container = 100 mL
The initial concentration of A2B3 is:
[A2B3] = moles / volume
[A2B3] = 0.01 mol / 0.1 L
[A2B3] = 0.1 mol/L
At equilibrium, the moles of B is given as 0.006 mol. We can assume that all the moles of A2B3 reacted to form A and B. Since the reaction stoichiometry is 2:3 for A2B3:B, we can calculate the moles of A:
moles of A = (moles of B) * (2/3)
moles of A = 0.006 mol * (2/3)
moles of A = 0.004 mol
Now, let's calculate the concentrations of A and B at equilibrium.
[A] = moles of A / volume
[A] = 0.004 mol / 0.1 L
[A] = 0.04 mol/L
[B] = moles of B / volume
[B] = 0.006 mol / 0.1 L
[B] = 0.06 mol/L
Now, substitute these concentrations into the equilibrium constant expression:
K = ([A]^2 * [B]^3) / [A2B3]
K = (0.04 mol/L)^2 * (0.06 mol/L)^3 / (0.1 mol/L)
K = 0.000096 mol^2/L^3
Therefore, the value of K is 0.000096 mol^2/L^3.
To find the value of K, we need to use the expression for the equilibrium constant (Kc) for the given reaction:
Kc = [A]^2 * [B]^3
where [A] represents the concentration of A(g) and [B] represents the concentration of B(g) at equilibrium.
First, we need to determine the concentrations of A(g) and B(g) at equilibrium.
Given that the initial moles of A2B3 is 0.01 mol and the volume of the container is 100 mL, we can calculate the concentration of A2B3:
initial concentration of A2B3 = (0.01 mol) / (0.1 L) = 0.1 M
Since the stoichiometry of the reaction tells us that for every 1 mol of A2B3, we get 3 mol of B(g), we can calculate the concentration of B(g) at equilibrium:
[B] = (0.006 mol) / (0.1 L) = 0.06 M
Substituting these values into the equilibrium constant expression, we have:
Kc = ([A]^2) * ([B]^3)
= (0.1 M)^2 * (0.06 M)^3
= 0.01 * 0.000216
= 0.00000216
Therefore, the value of K for the given reaction is 0.00000216.