A(aq) <--> 2B(aq)
Kc= 5.98 * 10^-6 at 500 K
If a 2.30 M sample of A is heated to 500 K, what is the concentration of B at Equilibrium?
To solve this problem, we can use the concept of the equilibrium constant and the stoichiometry of the balanced chemical equation. Here's how you can find the concentration of B at equilibrium:
1. Write the balanced chemical equation: A(aq) <--> 2B(aq)
2. Identify the equilibrium constant expression (Kc) for the reaction:
Kc = [B]^2 / [A]
3. Given that Kc = 5.98 * 10^-6, we can set up the equation:
5.98 * 10^-6 = [B]^2 / [A]
4. We know that the initial concentration of A is 2.30 M. Since the reaction has not yet reached equilibrium, we can assume that the concentration of B is 0 initially.
5. Let's denote the concentration of B at equilibrium as x. Therefore, the concentration of A at equilibrium will be (2.30 - x) (since A is being converted to B).
6. Substitute the values into the equation:
5.98 * 10^-6 = (x^2) / (2.30 - x)
7. Simplify the equation by multiplying both sides by (2.30 - x):
5.98 * 10^-6 * (2.30 - x) = x^2
8. Rearrange the equation to form a quadratic equation:
x^2 + (5.98 * 10^-6)(x) - (5.98 * 10^-6)(2.30) = 0
9. Solve the quadratic equation using the quadratic formula or a calculator. You will find two possible values for x since it's a quadratic: x = 0.0206 or x = -0.0206.
10. Since concentration cannot be negative, we discard the negative value. Therefore, the concentration of B at equilibrium is approximately 0.0206 M.
So, the concentration of B at equilibrium is approximately 0.0206 M.