Question

Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a tank with of ammonia gas, and when the mixture has come to equilibrium measures the amount of nitrogen gas to be 13. mol. Calculate the concentration equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.

Answer 1

Complete Question

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Answer:

The concentration equilibrium constant is $K_c = 14.39$

Explanation:

The chemical equation for this decomposition of ammonia is

                $2 NH_3$  ↔   $N_2 + 3 H_2$

The initial concentration of ammonia is mathematically represented a

          $[NH_3] = \frac{n_1}{V_1} = \frac{29}{75}$

          $[NH_3] = 0.387 \ M$

The initial concentration of nitrogen gas  is mathematically represented a

         $[N_2] = \frac{n_2}{V_2}$

         $[N_2] = 0.173 \ M$

So  looking at the equation

   Initially (Before reaction)

      $NH_3 = 0.387 \ M$

      $N_2 = 0 \ M$

      $H_2 = 0 \ M$

During reaction(this is gotten from the reaction equation )

        $NH_3 = -2 x$(this implies that it losses two moles of concentration )

         $N_2 = + x$  (this implies that it gains 1 moles)

         $H_2 = +3 x$(this implies that it gains 3 moles)

Note : x denotes concentration

At equilibrium

        $NH_3 = 0.387 -2x$

       $N_2 = x$

        $H_2 = 3 x$

Now since

     $[NH_3] = 0.387 \ M$

     $x= 0.387 \ M$    

$H_2 = 3 * 0.173$    

$H_2 = 0.519 \ M$    

$NH_3 = 0.387 -2(0.173)$

$NH_3 = 0.041 \ M$

Now the equilibrium constant is

           $K_c = \frac{[N_2][H_2]^3}{[NH_3]^2}$

substituting values

           $K_c = \frac{(0.173) (0.519)^3}{(0.041)^2}$

           $K_c = 14.39$