Question

1 mol of SO3 is placed in a reaction vessel..
2SO3 ------: 2SO2 +O2
at equilibrium 0.6 mol of SO2 were found..The equilibrium constant of the reaction will be ?​

Answer 1

Answer:

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Answer 2

The equilibrium constant of the reaction is 0.675

Explanation:

We are given:

Initial moles of $SO_3$ = 1.0 moles

Equilibrium moles of $SO_2$ = 0.6 moles

For the given chemical equation:

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

Initial:          1.0

At eqllm:     1-2x         2x     x

Evaluating the value of 'x'

$\Rightarow 2x=0.6\\\\\Rightarrow x=0.3$

The expression of $K_p$ for the above equation follows:

$K_{eq}=\frac{[SO_2]^2\times [O_2]}{[SO_3]}$

$[SO_3]=1-2x=[1-(2\times 0.3)]=0.4moles$

$[SO_2]=0.6moles$

$[O_2]=x=0.3moles$

Putting values in above equation, we get:

$K_{eq}=\frac{(0.6)^2\times 0.3}{(0.4)^2}\\\\K_{eq}=0.675$

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