Question

What is the concentration of NOBr, if the concentration of NO was measured to be 0.89 M, Br2 was measured to be 0.562 M, and the equilibrium constant, K, is 1.3 × 10-2?

Answer 1

Answer:

0.076M = [NOBr]

Explanation:

For the reaction:

2NO + Br₂ ⇄ 2 NOBr

The equilibirum constant, K, is defined as:

$K = \frac{[NOBr]^2}{[NO]^2[Br_2]}$(1)

Replacing the concentrations and the equilibrium value in (1):

$1.3x10^{-2} = \frac{[NOBr]^2}{[0.89]^2[0.562]}$

5.79x10⁻³ = [NOBr]²

0.076M = [NOBr]

I hope it helps!