If a 4.30 M sample of A is heated to 500 K, what is the concentration of B at equilibrium?
A(aq) --><--- 2B(aq) Kc=8.54 x 10^-6 at 500K
---><--- = equilibrium
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To find the concentration of B at equilibrium, we can use the equilibrium constant expression, Kc.
The equilibrium constant expression for the given reaction is:
Kc = [B]^2 / [A]
We know the value of Kc (8.54 x 10^-6) and the initial concentration of A (4.30 M). We need to determine the concentration of B at equilibrium.
Let's assume x mol/L of A is converted into B at equilibrium. As a result, 2x mol/L of B will be formed.
The concentration of A at equilibrium will be (4.30 - x) mol/L.
The concentration of B at equilibrium will be 2x mol/L.
Using these concentrations in the equilibrium constant expression, we can set up the equation:
Kc = (2x)^2 / (4.30 - x)
Simplifying the equation:
8.54 x 10^-6 = 4x^2 / (4.30 - x)
Cross-multiplying and rearranging the equation:
8.54 x 10^-6 * (4.30 - x) = 4x^2
Now we have a quadratic equation. We can multiply both sides by 4x and rewrite it in standard form:
34.16 x 10^-6 - x * 8.54 x 10^-6 = 4x^2
34.16 x 10^-6 - 8.54 x 10^-6 * x = 4x^2
4x^2 + (8.54 x 10^-6 * x) - 34.16 x 10^-6 = 0
Now, we can solve this quadratic equation to find the value of x. Once we have the value of x, we can calculate the concentration of B at equilibrium using 2x.