Phosphene decomposes to phosphorus and hydrogen in a first order mechanism. 4PH3(g) --> P4(g) + 6H2(g)

Question

This reaction's half-life is 35.0 s at 680 degrees C. If you had 520 mmHg of phosphene in an 8.00-L flask, how long does it take for the pressure in the flask to rise to standard pressure?

The answer is 48.3 s, I am just not sure how to do it out.

Answer 1

4PH3(g)==> P4(g) + 6H2(g)
For a 1st order rxn k = 0.693/t_1/2 = 0.693/35 = 0.0198

pPH3 decreases, pP4 increases, pH2 increases; we want the total to be 760 mm.
Therefore 520-4p+p+6p = 760 and p = 80 mm
So pPH3 must decrease from 520 to 520-4(80) = 200.
Then ln(No/N) = kt and
ln(520/200) = 0.0198t
Solve for t in seconds.

Answer 2

I can follow this until we get to the

520-4p+p+6p=760 and p=80 mm

The 520 is obviously the starting pressure and the 760 the final, but I'm not sure how you are getting the amounts that you are subtracting since they don't seem to match the equation to me. Could you please explain a bit more where the specific numbers come from?

Answer 3

Very Painful that this question was asked about to years ago ... And Answers is probably being dropped after 10 Years

To Solve this Question, we must first understand that pressure is Directly proportional to Concentration
I.e P ∝ [ ]

Using The Rate equation for the first order reaction, although it has many forms,
Ln [A]f - Ln [A]° = -KT
That is Naturally log of the final Concentration - Natural Log of the initial concentration...

Since Pressure is proportional we can have
Ln Pf - Ln P° = -KT......... (1)
Half life for first order reaction is

t½ = 0.693/K ........ (2)

We Can Find our K from this Half life equation in eqn (2)

K = 0.693/35
K 0.0198 per sec
The Starting pressure is 520mmHg
If the equation is
4PH3(g) --> P4(g) + 6H2(g)
It shows that the initial pressure of PH3 is 520, The Final Pressure of Phosphine left can't be up to 520mmHg as it's decomposing to give P4(g) and H2(g)

Here we might need a bit of the knowledge of Mole constituents and to make it easier let's work with 1 mole of PH3(g)

Dividing the equation through by 4,
We have
4PH3/4(g)--> P4/4(g) + 6H2/4(g)
PH3(g) ---> P4/4(g) + 6H2/4 (g)

After reaction of PH3, x pressure have been removed from PH3 and used to form x pressure of P4 and x pressure of H2, so what is left of PH3 is (520 - x) this is the final Pressure... Let's get back to the equation

PH3(g) ---> P4/4(g) + 6H2/4 (g)
(520-x) (x/4) + (6x/4)

Addition of all these pressures equals the total atmospheric pressure as specified in the question
I.e
(520-x) + (x/4) + (6x/4) = 760mmHg
If well solved our X = 320mmHg

What we need is the Value of (520-x), because that's the final pressure we need in our equation
520-320 = 200mmHg

Now that we have got our final Pressure we can now go back to equation (1) and insert it. From there we find the time

Ln 200 - Ln 520 = -0.0198T
-0.956 = -0.0198T

T = 48.28S

I wish you good luck