Question

Consider the following reaction:2NOBr(g) 2NO(g) + Br2(g)If 0.412 moles of NOBr(g), 0.678 moles of NO, and 0.224 moles of Br2 are at equilibrium in a 10.3 L container at 516 K, the value of the equilibrium constant, Kc, is:

Answer 1

Answer: $K_c=0.0587M$

Explanation: The given chemical reaction is:

$2NOBr(g)\rightleftharpoons 2NO(g)+Br_2(g)$

Equilibrium constant (Kc) in general is written as:

$K_c=\frac{[products]}{[reactants]}$

Note:- Coefficients are written as their powers

So, the Kc expression for the above reaction will be:

$K_c=\frac{[NO]^2[Br_2]}{[NOBr]^2}$

Equilibrium moles are given for all of them. Let's divide the moles by given liters to get the concentrations.

$[NOBr]=\frac{0.412mol}{10.3L}$   = 0.040 M

$[NO]=\frac{0.678mol}{10.3L}$   = 0.0658 M

$[Br_2]=\frac{0.224mol}{10.3L}$   = 0.0217 M

Plug in the values in the equilibrium expression to calculate Kc.

$K_c=\frac{(0.0658M)^2(0.0217M)}{(0.040M)^2}$

$K_c=0.0587M$