Question

A student ran the following reaction in the laboratory at 254 K:

2NO(g) + Br₂(g) ⇌  2NOBr(g)

When she introduced 0.185 moles of NO(g) and 0.130 moles of Br₂(g) into a 1.00 liter container, she found the equilibrium concentration of NOBr(g) to be 0.152 M.

Calculate the equilibrium constant, K_c, she obtained for this reaction.

K_c =

Answer 1

reaction is

2NO(g) + Br₂(g) ⇌  2NOBr(g)

E..> (0.185 - 2x) ...(0.130 - x) ..........+ 2x

2 mole No react with one mole Br2 to form 2 mole NOBr.

at  equilibrium,

[NOBr(g)] =  0.152 M

thus

2x = 0.152 M

or

x = 0.076 M

[NO] = (0.185 - 2x) = (0.185 - 2 * 0.076) = 0.033 M

and

[O2] = (0.130 - x) = (0.130 - 0.076) = 0.054 M

Kc = [NOBr]^2 / [NO]^2 [Br2]

or

Kc = (0.152)^2 / (0.033)^2 * 0.054 = 392.9

or

Kc = 393 (answer)