Initially, 0.600 mol of A is present in a 4.00-L solution.
2A[aq]->2B[aq]+C[aq]
At equilibrium, 0.200 mol of C is present. Calculate K.
Well, let's see if we can solve this equation without getting too tangled up!
The equation is 2A[aq] -> 2B[aq] + C[aq]. We can see that for every 2 moles of A, we get 2 moles of B and 1 mole of C.
At the start, we have 0.600 moles of A, so that means we'll get twice as many moles of B and the same number of moles of C.
That gives us 1.200 moles of B and 0.600 moles of C at the start.
Now, at equilibrium, we have 0.200 moles of C. Since we're getting the same number of moles of C at equilibrium as we started with, that means we've used up 0.400 moles of C from the initial amount.
So now we have 0.400 moles of C and 1.200 moles of B.
Now, we can calculate the equilibrium constant, K. Since K = [B]^2 [C] / [A]^2, we can plug in the numbers:
K = (1.200)^2 * (0.400) / (0.600)^2
K = 0.576
So the equilibrium constant, K, is 0.576.
Hope that didn't leave you feeling too twisted up like a pretzel!
To calculate the equilibrium constant, K, for the given reaction, we need to use the molar concentrations of the reactants and products at equilibrium.
Given:
Initial moles of A = 0.600 mol
Equilibrium moles of C = 0.200 mol
We can convert the moles to molar concentrations by dividing by the volume of the solution:
Initial concentration of A = (0.600 mol) / (4.00 L)
Equilibrium concentration of C = (0.200 mol) / (4.00 L)
The stoichiometric coefficients in the balanced equation tell us that the concentration of B is also equal to the concentration of C, but since we are only given the concentration of C, we will assume it represents the concentration of B as well.
Therefore, the equilibrium concentrations of B and C are both equal to (0.200 mol) / (4.00 L).
Now we can write the expression for the equilibrium constant, K:
K = ([B]^2 * [C]) / ([A]^2)
Substituting the equilibrium concentrations we calculated:
K = ((0.200 mol / 4.00 L)^2 * (0.200 mol / 4.00 L)) / ((0.600 mol / 4.00 L)^2)
K = (0.0125 M^3) / (0.0375 M^2)
K = 0.33333
Therefore, the equilibrium constant, K, is approximately 0.33333.
To calculate the equilibrium constant (K) for this reaction, you need to know the concentrations of each species involved at equilibrium. In this case, you have the initial concentration of A and the equilibrium concentration of C.
However, in order to calculate the equilibrium concentrations of A, B, and C, you need additional information, specifically the stoichiometric coefficients and the changes that occur in each species during the reaction. With this information, you can set up an equilibrium expression and solve for K.
From the balanced equation:
2A(aq) → 2B(aq) + C(aq)
We know that for every 2 moles of A that react, you will form 2 moles of B and 1 mole of C.
Let's define "x" as the change in concentration of A. Since 2 moles of A give rise to 1 mole of C, the change in concentration of C will be "x/2".
At equilibrium, the concentration of A will be its initial concentration minus the change: [A]eq = [A]initial - x
The concentration of B at equilibrium will be 2 times the change: [B]eq = 2x
The concentration of C at equilibrium will be the change divided by 2: [C]eq = x/2
Since K is calculated using concentrations, we can now write the expression for K:
K = ([B]eq)^2 * ([C]eq) / ([A]eq)^2
Substituting the equilibrium concentrations we found earlier:
K = (2x)^2 * (x/2) / ([A]initial - x)^2
Given that [A]initial = 0.600 mol and [C]eq = 0.200 mol, we can now solve for K.
K = (2x)^2 * (x/2) / (0.600 - x)^2
To find x, we can use the fact that at equilibrium the moles of A that have reacted will be equal to the moles of C that have formed:
0.600 - x = x
Solving for x:
2x = 0.600
x = 0.300
Now, substitute x = 0.300 into the expression for K:
K = (2 * 0.300)^2 * (0.300/2) / (0.600 - 0.300)^2
K = 0.6^2 * 0.150 / 0.3^2
K = 0.36 * 0.150 / 0.09
K = 0.054 / 0.09
K = 0.6
Therefore, the equilibrium constant (K) for this reaction is 0.6.
K= [B]^2*[C}/[A]^2
A=(.6-x)/4
B= (x)
C=(x/2)=.2
use C to solve for x, then solve for the rest.