Consider the below reaction. � �(𝑔)+3𝐵(𝑔)⇌2𝐶(𝑔) If an equi-molar mixture of A(g) and B(g) is kept inside a container at constant temperature, what is the decrease of pressure in the container when 10% of A(g) is reacted with B(g)? A) 5% B) 8% C) 10% D) 12% E) 15%

To solve this problem, we need to understand the stoichiometry of the reaction.

The reaction is: A(g) + 3B(g) ⇌ 2C(g)

This means that for every one mole of A(g) that reacts, three moles of B(g) also react, and two moles of C(g) are produced.

If 10% of A(g) is reacted, then 10% of B(g) will also react, because they are in an equi-molar mixture. This means that the total number of moles of A(g) and B(g) that have reacted is 10% + 10% = 20%.

However, for every one mole of A(g) that reacts, two moles of C(g) are produced. So, if 10% of A(g) has reacted, then 20% of C(g) has been produced.

So, the total change in the number of moles (and therefore the total change in pressure, since pressure is proportional to the number of moles for a gas at constant temperature and volume) is the sum of the decrease due to A(g) and B(g) reacting and the increase due to C(g) being produced.

This is 20% (decrease) - 20% (increase) = 0%.

So, there is no decrease in pressure in the container when 10% of A(g) is reacted with B(g). None of the options A) 5%, B) 8%, C) 10%, D) 12%, E) 15% are correct.