Consider the reaction A\(_{(s)}\) + 2B\(_{(g)}\) → 2C\(_{(aq)}\) + D\(_{(g)}\) What will be the effect...

Consider the reaction
A\(_{(s)}\) + 2B\(_{(g)}\) → 2C\(_{(aq)}\) + D\(_{(g)}\)
What will be the effect of a decrease in pressure on the reaction?

Given: The equation below
A\(_{(s)}\) + 2B\(_{(g)}\) → 2C\(_{(aq)}\) + D\(_{(g)}\)

According to Le Chatelier's Principle:

Pressure changes affects only the gaseous reactants and products
There must be different amount of gaseous particles on both sides.

Generally; a decrease in the pressure shifts the equilibrium position in favor of the side with larger volume of gases.

I.e rates of backward reaction.

NB: This is a correction to the video explanation.

There is an explanation video available below.

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