## Homework Statement

Consider a first order decomposition reaction

2P(g) -----> 4Q(g) + R(g) + S(l)

taking place at a temperature T. After 30 minutes from the start of the decomposition in a closed vessel, the total pressure developed is found to be 317 mmHg and after a long time, the total pressure observed was 617 mmHg. Calculate the toal pressure in the vessel after 75 minutes. (Given : V.P. of S(l) at temperature T=32.5 mmHg)

Attempt

Let the initial pressure of P be Pi.
At t=30 minutes, let 2x be the decrease in pressure of P.
There will be a pressure of 4x,x, ? due to Q,R,S respectively.

I have a problem in finding out the pressure of S. How is the partial pressure of S in the vessel related to the state of S? i.e. How do I know whether S will remain a liquid or change into gas? Someone suggested me that if the pressure due to S is say 'y', it will remain a liquid if y > 32.5mmHg.

I don't understand the concept behind it. Can somebody explain?

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Borek
Mentor
Liquid S doesn't appear till S pressure is not 32.5 mmHg. After that S pressure is constant.

I did not get you completely.
If partial pressure of S during the course of the reaction is less than 32.5 mmHg, then S will be a gas? Why is it so?
If partial pressure of S is greater than 32.5 mmHg, then why is it constant? Does it still remain a gas?

Borek
Mentor
By definition of saturated vapor - it is one that is in equilibrium with liquid. When pressure is lower - there is no liquid. When the pressure is higher - gas condenses till the pressure falls to the saturated.

Thanks for that Mr.Ph. I have another question -
Why do we consider the partial pressure of S for its state rather than the total pressure in the vessel which is always greater than 32.5 mmHg?

Borek
Mentor

Before partial pressure of S goes up to 32.5 we treat it exactly as every other substance, when it starts to condense we no longer can.

I meant that why does the state of S (whether liquid or gas) depends on its own partial pressure and not on the total pressure in the vessel (in which the reaction is carried out).

It is given that the pressure in the vessel is 317 mmHg after 30 minutes. Can't we just say that S will remain as liquid at t=30 minutes because 317 mmHg is greater than 32.5 mmHg?

Borek
Mentor
I meant that why does the state of S (whether liquid or gas) depends on its own partial pressure and not on the total pressure in the vessel (in which the reaction is carried out).
Because S doesn't care about other substances present and it is in equilibrium with its own vapor. That's the way it is, that's why partial pressures are that important.

You solved my BIG problem. Thank you very much.

I got one more problem When we heat water, it starts boiling only when its vapour pressure equals the atmospheric pressure which constitutes the pressure due to Nitrogen, Oxygen etc.
Why does its boiling point not depend only on the partial pressure of water vapour in air ?

Borek
Mentor
http://en.wikipedia.org/wiki/Boiling

While below the boiling point a liquid evaporates from its surface, at the boiling point vapor bubbles come from the bulk of the liquid. For this to be possible, the vapor pressure must be sufficiently high to win the atmospheric pressure

In our case S exists as a gas if its partial pressure is less than 32.5mmHg. Is it really a gas or vapors of the liquid?

Vapour is a Substance which is near the condensation but gas is very far away from condensation.In other words,gas is something that u cannot make it to liquid Only by increasing the pressure on it!! U have to decrease the tempreture aswel, otherwise it wouldnt turn into liquid!

there is a diff. b/w gas and vapour... 