Question

He equilibrium constant (KP) is 0.21 at a particular temperature for the reaction:

Answer 1

Explanation:

We will use the following equation,  

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

(a) Now, we will calculate the value of $Q_{p}$ as follows.

          $Q_{p} = \frac{(0.225)^{2}}{0.134}$

                      = 0.377

As, the $K_{p}$ is 0.21 (which is smaller than the value) this means that equilibrium will shift to the left .

(b)   $Q_{p} = \frac{(0.086)^{2}}{0.064}$

                  = 0.1155

Here, $Q_{p}$ is smaller than $k_{p}$ which means that the equilibrium will shift to the right.

(c)   $Q_{p} = \frac{(0.13)^{2}}{0.142}$

                   = 0.119

Here, $Q_{p}$ is smaller than $k_{p}$ which means that the equilibrium will shift to the right .

(d)    $Q_{p} = \frac{(0.173)^{2}}{0.142}$

                  = 0.21  

Since, here the value of $Q_{p}$ comes out to be equal to the given value. Therefore, this reaction is at equilibrium.

(e)  $Q_{p} = \frac{(0.151)^{2}}{0.062}$

                  = 0.36

Here, $Q_{p}$ is higher than $k_{p}$ the reaction will shift to the left.