Question

P(g) = Q(g) +R(g)
Q(g) +S(g) = R(g) +T(g)
P(g) +T(g) = R(g) +S(g)
Initially 1 mole each of P(g) and S(g) are present
5
while at equilibrium, [Q] = [T] and [S]
Mand
6
[R] = 1 M. The value of [S] is found to be x times
[P] at equilibrium. Find (10x).​

1

Answer 1

Answer:

$\huge\green\star\mathfrak{\underline{\underline{answer}}}$

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.

Answer 2

Answer:

sorry I didn't know.........