Question

Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 500. mL flask with 2.3 atm of ammonia gas at 32. °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of hydrogen gas to be 0.69 atm. Calculate the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.

Answer 1

Explanation:

Chemical reaction equation for the give decomposition of $NH_{3}$ is as follows:.

          $2NH_{3}(g) \rightleftharpoons N_{2}(g) + 3H_{2}(g)$

And, initially only $NH_{3}$ is present.

The given data is as follows.

  $P_{NH_{3}}$ = 2.3 atm at equilibrium

   $P_{H_{2}} = 3 \times P_{N_{2}}$ = 0.69 atm

Therefore,

          $P_{N_{2}} = \frac{0.69 atm}{3}$

                        = 0.23 aatm

So, $P_{NH_{3}}$ = 2.3 - 2(0.23)

                       = 1.84 atm

Now, expression for $K_{p}$ will be as follows.

         $K_{p} = \frac{(P_{N_{2}})(P^{3}_{H_{2}})}{(P^{2}_{NH_{3}})}$

           $K_{p} = \frac{(0.23) \times (0.69)^{3}}{(1.84)^{2}}$

                      = $\frac{0.23 \times 0.33}{3.39}$

                     = 0.0224

or,           $K_{p} = 2.2 \times 10^{-2}$

Thus, we can conclude that  the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture is $2.2 \times 10^{-2}$.