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Temka [501]
3 years ago
12

The decomposition of N2O4 into NO2 has Kp = 2. Some N2O4 is placed into an empty container, and the partial pressure of NO2 at e

quilibrium 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
Chemistry
1 answer:
Sunny_sXe [5.5K]3 years ago
6 0

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.

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