Question

A reaction mixture initially contains 0.86 atm NO and 0.86 atm SO3. Determine the equilibrium pressure of NO2 if Kp for the reaction at this temperature is 0.0118. NO(g) SO3(g) NO2(g) SO2(g)

Answer 1

Explanation:

Reaction equation for this reaction is as follows.

     $NO(g) + SO_{3}(g) \rightarrow NO_{2}(g) + SO_{2}(g)$

It is given that $K_{p}$ = 0.0118.

According to the ICE table,

              $NO(g) + SO_{3}(g) \rightarrow NO_{2}(g) + SO_{2}(g)$

Initial:           0.86      0.86               0            0

Change:          -x          -x                 +x           +x

Equilibrium:  0.86 - x   0.86 - x         x           x

Hence, value of $K_{p}$ will be calculated as follows.

           $K_{p} = \frac{P_{NO_{2}} \times P_{SO_{3}}}{P_{NO} \times P_{SO_{3}}}$

         0.0118 = $\frac{x \times x}{(0.86 - x)^{2}}$

             x = 0.084 atm

Thus, we can conclude that $P_{NO_{2}}$ is 0.084 atm.