AP Chemistry problem!?

Question

I came across this question from the 2000 AP exam, and it seemed easy and so I answered it by now I don't know if it's right!!

H2S(g) 2 H2(g) + S2(g)
(a reverse reaction!)

When heated, hydrogen sulfide gas decomposes according to the equation above. A 3.40 g sample of H2S(g) is introduced into an
evacuated rigid 1.25 L container. The sealed container is heated to 483 K, and 3.72 ×10-2 mol of S2(g) is present at equilibrium.

(a) Calculate the equilibrium concentration, in mol L-1, of the following gases in the container at 483 K.
(i) H2(g)
(ii) H2S(g)

(b) Calculate the partial pressure of S2(g) in the container at equilibrium at 483 K.


Thank you!

Answer 1

let us see the first part of the question.
3.4 g of H2S means 0.1 mole . this is the initial concentration of the H2S

THE REACTION: 2H2S------> 2H2 + S2
INITIAL MOLES: 0.1mole 0 mole 0 mole
equilibrium moles: 0.1-2x 2xmoles x moles
[as we know x=0.0372]
EQUILIBRIUM MOLES OF H2S=0.1-2(0.0372) =0.9256
EQUILIBRIUM MOLES OF H2=2(0.0372) =0.0744
EQUILIBRIUM MOLES OF S21=0.0372


equilibrium conc Of H2S=: 0.9256/1.25=0.74048mol L-1
equilibrium conc Of H2=0.0744/1.25 =0.05952molL-1
equilibrium conc Of S2=0.0372/1.25


So the equilibrium concentrations of H2S is 0.74048 mol L-1

AND H2 is 0.05952 mol l-1

Now let us see the second part of the question:
partial pressure of the S2= MOLE FRACTION OF S2 X TOTAL PRESSURE OF THE EQUILIBRIUM MIXTURE

But there is no equilibrium pressure in the problem????????????? If you miss it then substitute in the above formula. you will get the answer.

Answer 2

2000 Ap Chemistry Free Response Answers

Answer 3

artwork out the molecular weights for CH4 and a pair of O2 note that they occupy an section (quantity) at 2 atms even as the reaction takes position there is in common words one gas last and also you pick to understand the molecular weight of this gas analyze Avagadros huge variety (to get an information) and use P1V1/T1 = P2V2/T2 or PV=RT (T is in Kelvin)

Answer 4

(a)(i) The amount of H2 is also 3.72x10^-2 mol from the balanced equation. (3.72mol)/(1.25L) = 2.98x10^-2 mol/L

(a)(ii) (3.72x10^-2molS2) x (2molH2S)/1molS2) = 7.44x10^-2 molH2S consumed.

3.4gH2S = 0.1molH2S = 10x10^-2molH2S originally.

10X10^-2 - 7.44x10^-2 = 2.56x10^-2molH2S left.

(2.56x10^-2)/(1.25L) = 2.05x10^-2 mol/L H2S

(b) PV = nRT In this case, R = 0.0821 L-atm/mol-K or

R = 62.3 L-mmHg/mol-K mmHg is also called torr.