If a 2.30 M sample of A is heated to 500 K, what is the concentration of B at equilibrium?
A<-> 2B
Kc= 5.04x10^-6 at 500k
.........A <==> 2B
I......2.30M.....0
C........-x......2x
E......2.30-x....2x
Kc = 5.04E-6 = (B)^2/A
Substitute the E line into Kc expression I have above and solve for x.
Then 2x = (B)
Thanks!
To determine the concentration of B at equilibrium, we need to use the equilibrium constant (Kc) and the initial concentration of A.
Given:
Initial concentration of A (A0) = 2.30 M
Equilibrium constant at 500 K: Kc = 5.04 x 10^(-6)
Since A reacts to form 2B, the concentration of B at equilibrium will depend on the decrease in the concentration of A.
Let's assume the change in concentration of A is "x" M. At equilibrium, the concentration of A will be (A0 - x) M.
According to the balanced equation: A -> 2B, the concentration of B formed will be 2x M.
Using the equilibrium expression for the reaction: Kc = [B]^2/[A]
Substituting the values:
Kc = (2x)^2/(A0 - x)
Rearranging the equation:
4x^2 = Kc * (A0 - x)
Plugging in the given values:
4x^2 = (5.04 x 10^(-6)) * (2.30 - x)
Now we need to solve this quadratic equation to find the value of x, which represents the change in concentration of A:
4x^2 = (5.04 x 10^(-6))(2.30 - x)
Expanding and rearranging the equation:
4x^2 = 11.512 x 10^(-6) - 5.04 x 10^(-6) * x
Bringing all terms to one side:
4x^2 + 5.04 x 10^(-6) * x - 11.512 x 10^(-6) = 0
This equation can be solved using the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / 2a
Using the coefficients for our equation:
a = 4
b = 5.04 x 10^(-6)
c = -11.512 x 10^(-6)
Solving the quadratic equation will give us the value(s) for x. By substituting the value of x into (A0 - x) and 2x, we can calculate the equilibrium concentrations of A and B.
To find the concentration of B at equilibrium, we can use the equilibrium constant expression and the initial concentration of A.
The equilibrium constant expression for the given reaction is:
Kc = [B]^2 / [A]
In this case, we are given Kc = 5.04 x 10^-6 and the initial concentration of A is 2.30 M.
Since the stoichiometric coefficient of A in the balanced equation is 1, the initial concentration of A remains unchanged throughout the reaction. Therefore, [A] at equilibrium is also 2.30 M.
Let's represent the concentration of B at equilibrium as x M. Since two moles of B are produced for every mole of A that reacts, the concentration of B at equilibrium is 2x M.
Now, we can substitute the known values into the equilibrium constant expression:
5.04 x 10^-6 = (2x)^2 / 2.30
Simplifying the equation:
5.04 x 10^-6 = 4x^2 / 2.30
Multiply both sides of the equation by 2.30 to eliminate the denominator:
11.592 x 10^-6 = 4x^2
Divide both sides by 4:
2.898 x 10^-6 = x^2
Take the square root of both sides:
x = √(2.898 x 10^-6)
Calculating the value of x, we find:
x ≈ 0.0017 M
Therefore, the concentration of B at equilibrium is approximately 0.0017 M.