Question

Two moles of compound P were placed in a vessel. The compound P was partly decomposed by
heating. A dynamic equilibrium between chemicals P, Q and R was established.
At equilibrium, x mol of R were present and the total number of moles present was (2 + x).
What is the equation for this equilibrium?
A P ---->< 2Q + R
B 2P ----><2Q + R
C 2P----->< Q + R
D 2P ---->< Q + 2R

Answer 1

hm i think it has to be B, you started off with 2 moles of P

if 1 mole of P decomposed to 1 mole of Q and x mole of R you'd be left with 2+x

otherwise you would be left with other values. For example, if we consider A

1 mole of P decomposed to 2 moles of Q and x moles of R, then you'd have 3+x moles

you can invalidate C and D with the same rationale.

Answer 2

in B the 2 moles of p are decomposed to 2 moles of Q and x moles of R is that what you meant

Answer 3

no, it says P partially decomposes, meaning there is some left

Answer 4

in C P decomposed to 1 mole and 1 mol of Q so we have a total of 2x+2 am i right.

Answer 5

In B 2 moles are decomposed to 2x of Q and an x of P

Answer 6

in C, if 1 mole of P decomposed, you'd be left with 1 mole of P, 1/2 mole of C and x moles of R

in B, if both moles of P decomposed, you'd be left with 2 moles of Q and x moles of R (1 mole of R, really).

Answer 7

and B is reduced to 2-x

Answer 8

of P

Answer 9

i'm not sure i follow your last post

Answer 10

correction:

in C, if 1 mole of P decomposed, you'd be left with 1 mole of P, 1/2 mole of \(\color{red}Q\) and x moles of R

Answer 11

thank you for your help but in C 2 moles were present so only 1 mole of Q is present and x moles of R

Answer 12

so why is it 1/2 mole of Q in C

Answer 13

no problem.

nope, look at the ratio,

\(\dfrac{n_P}{2}=\dfrac{n_Q}{1}\)

1 mole of P decomposed:

\(\dfrac{1}{2}=\dfrac{n_Q}{1}\rightarrow n_Q=\dfrac{1*1}{2}=\dfrac{1}{2}\)

so you made 1/2 a mole of Q

Answer 14

okay is this right for option C