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=([products])/([reactants])`

Note:- Coefficients are written as their powers

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

`K_c=([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]=(0.412mol)/(10.3L)` = 0.040 M

`[NO]=(0.678mol)/(10.3L)` = 0.0658 M

`[Br_2]=(0.224mol)/(10.3L)` = 0.0217 M

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

`K_c=((0.0658M)^2(0.0217M))/((0.040M)^2)`

`K_c=0.0587M`