Answer: No, the reverse reaction must proceed to establish equilibrium.
Explanation:
[tex]K_p[/tex] is the constant of a certain reaction at equilibrium while [tex]Q_p[/tex] is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.
For the given chemical reaction:
[tex]H_2(g)+I_2(g)\rightleftharpoons 2HI(g)[/tex]
The expression of [tex]Q_p[/tex] for above equation follows:
[tex]Q_p=\frac{(p_{HI})^2}{p_{H_2}\times p_{I_2}}[/tex]
We are given:
[tex]p_{HI}=1.055atm\\p_{H_2}=0.127atm\\p_{I_2}=0.134atm[/tex]
Putting values in above equation, we get:
[tex]Q_p=\frac{(1.055)^2}{0.127\times 0.134}=65.41[/tex]
We are given:
[tex]K_p=55.2[/tex]
There are 3 conditions:
As, [tex]Q_p>K_p[/tex], the reaction will be favoring reactant side or the reaction must proceed in the reverse direction.
Hence, no, the reverse reaction must proceed to establish equilibrium.