Question

A 100 L reaction container is charged with 0.571 mol of NOBr, which decomposes at a certain temperature** (say between 100 and 150 ^oC) according to the following reaction:

NOBr(g) ? NO(g) + 0.5Br₂(g)

At equilibrium the bromine concentration is 1.28x10^-3 M. Calculate K_c (in M^0.5)

**Not specifying the temperature allows for a more liberal use of random numbers.

Answer 1

initial [NOBr] = 0.571 moles/100 L = 5.71 x 10^-3 M

at equilibrium,

[Br2] = 1.28 x 10^-3 M

From the reaction,

2NOBr <==> 2NO + Br2

equilibrium concentration of,

[NOBr] = 5.71 x 10^-3 - 2 x 1.28 x 10^-3 = 3.15 x 10^-3 M

[NO] = 2 x 1.28 x 10^-3 = 2.56 x 10^-3 M

So,

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

     = (2.56 x 10^-3)^2 x (1.28 x 10^-3)/(3.15 x 10^-3)^2

     = 8.45 x 10^-4

in terms of M^0.5

Kc = sq.rt(8.45 x 10^-4) = 2.91 x 10^-2