Question

Assume that the change in concentration of N2O4 is small enough to be neglected in the following problem. Calculate the equilibrium concentration of both species in 1.00 L of a solution prepared from 0.129 mol of N2O4 with chloroform as the solvent.

Answer 1

The question is incomplete, here is the complete question:

Assume that the change in concentration of N₂O₄ is small enough to be neglected in the following problem. Calculate the equilibrium concentration of both species in 1.00 L of a solution prepared from 0.129 mol of N₂O₄ with chloroform as the solvent.

N₂O₄ ⇆ 2NO₂ ;   Kc = 1.07×10⁻⁵  in chloroform

Answer: The equilibrium concentration of $N_2O_4$ and $NO_2$ are 0.1233 M and 0.0114 M respectively

Explanation:

We are given:

Initial moles of $N_2O_4$ = 0.129 moles

Volume of the solution = 1.00 L

The molarity is calculated by using the equation:

$\text{Molarity of the solution}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}$

Initial concentration of $N_2O_4=\frac{0.129}{1.00}=0.129M$

For the given chemical equation:

                  $N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initial:          0.129

At eqllm:     0.129-x        2x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

We are given:

$K_c=1.07\times 10^{-5}$

Putting values in above expression, we get:

$1.07\times 10^{-3}=\frac{(2x)^2}{(0.129-x)}\\\\x=-0.006,0.00574$

Neglecting the value of x = -0.006 because concentration cannot be negative

So, equilibrium concentration of $N_2O_4=(0.129-x)=(0.129-0.0057)=0.1233M$

Equilibrium concentration of $NO_2=2x=(2\times 0.0057)=0.0114M$

Hence, the equilibrium concentration of $N_2O_4$ and $NO_2$ are 0.1233 M and 0.0114 M respectively