Question

The decomposition of N2O4 into NO2 has Kp = 2. Some N2O4 is placed into an empty container, and the partial pressure of NO2 at equilibrium is measured to be 0.2 atm. What was the initial pressure in the container prior to decomposition? A) 0.10 atm B) 0.12 atm C) 0.20 atm D) 0.22 atm E) 0.30 atm

Answer 1

Answer: Option (B) is the correct answer.

Explanation:

Expression for the given decomposition reaction is as follows.

           $N_{2}O_{4} \rightarrow 2NO_{2}$

Let us assume that x concentration of $N_{2}O_{4}$ is present at the initial stage. Therefore, according to the ICE table,

                    $N_{2}O_{4} \rightarrow 2NO_{2}$

Initial :               x                   0

Change :       - 0.1        $2 \times 0.1$

Equilibrium : (x - 0.1)             0.2

Now, expression for $K_{p}$ of this reaction is as follows.

     $K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}$

Putting the given values into the above formula as follows.

          $K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}$

                 $2 = \frac{(0.2)^{2}}{(x - 0.1)}$

                $2 \times (x - 0.1) = (0.2)^{2}$

                            x = 0.12

This means that $P_{N_{2}O_{4}}$ = x = 0.12 atm.

Thus, we can conclude that the initial pressure in the container prior to decomposition is 0.12 atm.