The Kingdom of Chemithia- Part 5

Chemistry Level 4

You enter the lab for the Royal Chemists and are immediately stunned by how incredible it is. Every chemical imaginable is on the shelves and the apparatus is gleaming. You excitedly look around and go to the table to look closer. This is going to be a fun job.

As you keep testing chemicals and solving issues in the kingdom, one day, a royal messenger comes to your lab, saying that King Antimony requests your presence. You follow him to the lavish court, where King Antimony sits on his electrum throne.

"Greetings, Aurous. I have a problem that I would like you to solve." he says, "For some experiments, we need to be able to convert White Phosphorus into its vapours. The reaction takes place as: "

P 4 ( g ) ⇌ 2 P 2 ( g ) P_4 (g) ⇌ 2P_2 (g)

"So we are currently converting the White Phosphorus in a furnace at 1000 K 1000K . Our mixture that is formed has a total pressure of 16 a t m o s p h e r e s 16 atmospheres and a density of 18.6 g r a m s / l i t r e 18.6 grams/litre . What we want to find out is the value of K p K_p for this reaction, to know the ratio of pressures of the vapours and the phosphorus itself. Knowing this will help us to determine whether it is a viable method for the future or not."

What is the value of K p K_p for the above reaction?

[Useful info: take R R = 0.08 a t m − l i t r e / K − m o l e 0.08 atm-litre/K-mole ] [Assume all conditions are ideal]

This question is part of my set Tales of Chemithia ! Give it a try; it's a story woven together by Chemistry problems!


The answer is 8.00.

Jay Steller by Jay Steller , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Jay Steller
Mar 13, 2017

Solution:

  1. P M = d R T PM=dRT [Ideal Gas Law]
  2. M = 18.6 ∗ 0.08 ∗ 1000 / 16 M=18.6*0.08*1000/16 . Therefore, M = 93. M=93.
  3. Initially there was 1 m o l e 1 mole .
  4. P 4 ⇌ 2 P 2 P_4 ⇌ 2P_2 . Suppose x x moles are consumed in reaction.
  5. Equilibrium concentration of P 4 P_4 : 1 − x 1-x
  6. Equilibrium concentration of P 2 P_2 : 2 x 2x
  7. M a v g = 93 = [ ( 1 − x ) ∗ 124 ] + [ 2 x ∗ 62 ] / ( 1 − x + 2 x ) M_{avg}=93=[(1-x)*124]+[2x*62]/(1-x+2x) . Simplifying, we get x = 0.5 x=0.5 .
  8. Therefore molar ratio P 4 : P 2 P_4:P_2 = ( 2 / 3 ) : ( 2 / 3 ) (2/3):(2/3) = 1 : 1 1:1
  9. If total pressure of system is 16 a t m 16 atm , P r e s s u r e P 4 = 8 a t m Pressure_{P_4}=8 atm and P r e s s u r e P 2 = 8 a t m Pressure_{P_2}= 8 atm
  10. Hence K p = 8 2 / 8 K_p=8^2/8 . The answer is 8.
5 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...