Question

In the reaction 2P(g) + Q(g) = 3R(g) + S(g). If
2 mol each of P and Q taken initially in a
1 L flask. At equilibrium which is true:-
(1) [P] < [Q]
(2) [P] = [Q]
(3) [Q] = [R]
(4) None of these​

Answer 1

Answer:

Option (1) [P] < [Q]

Explanation:

In the reaction 2 moles of P reacts with 1 mole of Q to give 3 moles of R and 1 mole of S

$2P+Q \rightarrow 3R+S$

If [R] = 3x and [S] = x

Then [P]=2-2x and [Q]=2-x

Thus at equilibrium the concentration of Q will be more than that of P

Thus

[P] < [Q]

Hope this helps.

Answer 2

ANSWER

2P

(g)

​

+Q

(g)

​

⇌3R

(g)

​

+S

(g)

​

Initially 2moles 2moles 0 0

At eqm 2−2α 2−α α α

Here 2 moles of P reacts with 1 mole of Q. So at equilibrium number of moles of P consumed will be higher than that of Q. Therefore, rate of decrease of [P] is higher than that of [Q].

∴[P]<[Q] at equilibrium.