Question

Consider the reaction below 2SO2(g) + O2(g) ⇌ 2SO3(g) At 1000 K the equilibrium pressures of the three gases in one mixture were found to be 0.562 atm SO2, 0.101 atm O2, and 0.332 atm SO3. Calculate the value of Kp for the reaction.

Answer 1

Answer:$K_p$ for the reaction is 3.45

Explanation:

The balanced chemical reaction is:

          $2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$                  

At eqm.   0.562 atm   0.101 atm    0.332 atm

As we are given that:

The expression of $K_p$ for above equation follows:

$K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times p_{O_2}}$

Putting values in above equation, we get:

$K_p=\frac{(0.332)^2}{(0.562)^2\times 0.101}=3.45$

The value of $K_p$ for the reaction is 3.45