what is Kc for 2NOBr-->2NO+Br2 when the temperature is 1000K and the Kc is 1.3x10^-2

What's the temperature for the Kc listed?

To find the value of the equilibrium constant (Kc) for the reaction 2NOBr ⇌ 2NO + Br2 at a temperature of 1000K, we can use the given value of Kc. Here's how you can calculate it:

1. Write down the balanced chemical equation:
2NOBr ⇌ 2NO + Br2

2. The equilibrium constant expression (Kc) for this reaction is:
Kc = [NO]^2 * [Br2] / [NOBr]^2
(Note: The coefficient of each species in the balanced equation becomes its exponent in the Kc expression.)

3. Substitute the given Kc value into the equation:
1.3x10^-2 = [NO]^2 * [Br2] / [NOBr]^2

4. Since we don't know the individual concentrations of NO, NOBr, and Br2, we can use the concept of stoichiometry to simplify the equation further.

5. Imagine starting with an initial concentration of NOBr and letting the reaction proceed to reach equilibrium. In this case, let's assume that the initial concentration of NOBr is "x". Since each NOBr molecule gives rise to 2 NO molecules and 1 Br2 molecule, at equilibrium, the concentration of NO will be 2x, and the concentration of Br2 will be x.

6. Substitute the equilibrium concentrations into the Kc expression:
1.3x10^-2 = (2x)^2 * (x) / (x)^2
Note: The concentration of NOBr cancels out because it is raised to the power of 2 in the Kc expression numerator and denominator.

7. Simplify the equation further:
1.3x10^-2 = 4x^3 / x^2
(Note: When multiplying with the same base, we add their exponents, hence 2*x^2)

8. Rearrange the equation:
1.3x10^-2 = 4x
Divide both sides by 4:
3.25x10^-3 = x

9. The concentration of NOBr at equilibrium (x) is equal to 3.25x10^-3 M.

Therefore, the equilibrium constant (Kc) for the reaction 2NOBr ⇌ 2NO + Br2 at a temperature of 1000K and a Kc value of 1.3x10^-2 is approximately 3.25x10^-3.