Consider the reaction
A\(_{(s)}\) + 2B\(_{(g)}\) → 2C\(_{(aq)}\) + D\(_{(g)}\)
What will be the effect of a decrease in pressure on the reaction?
the equilibrium will shift
rate of forward reaction will increase
rate of backward reaction will increase
rates of forward and backward reaction are not affected
Explanation
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The Correct Answer Is C.
When there is an increase in pressure, the equilibrium will shift towards the side of the reaction with fewer moles of gas. When there is a decrease in pressure, the equilibrium will shift towards the side of the reaction with more moles of gas.
The answer is supposed to be C because pressure only affects gases. So it's only if the number of moles of the gases at both sides are equal, that's when an increase or decrease in pressure won't affect the equilibrium position. But in this scenario, a decrease in pressure on the reaction will favour the side that has an higher pressure (or higher number of moles) which is the backward reaction. So the answer should be C
Only gaseous molecules should be considered when pressure is involved but the answer u guys picked considered the solid and aqueous molecules which is very wrong
The effect of pressure affect only the gas compouds or element then according to the question a decrease in pressure will favour the backward formation leading to higher yield in the reactant because it has more number of moles in the gaseous element 😤
Change in pressure only affects a reaction in equilibrium when all the components are in the gaseous phase. In that particular reaction the components are in different phases.
Absolutely correct
in chemical reaction related to pressure
only gases are used
please the is wrong pressure only affects gases also they are equal numbers of mole on the reactant and product so there no effect
The correct answer is D.
Pressure only affects gases, True!
But this is not equilibrium constant that you cancel out the solid.
Kindly confirm and make necessary changes.
Thank you.
