[Marcus Soutlo]

Is my solution to this problem correct?

Reaction
Volume of 6% H2O2 (mL):10.0
Volume of distilled water (mL):0.0
Volume of 1.0 M NaI (mL):10.0
Volume of distilled water (mL):0.0   
O2 production in 60 s (mL):151.0

Question:
Calculate the number of mols of oxygen gas produced per second. Remember PV=nRT and that you must account for the partial pressure of water vapour.
The reactions conditions are: 
Atmospheric pressure: 102.6 kPa;
Temperature:  290 K, 
Partial pressure of water vapour:  1.94 kPa.

My answer:
Ptotal = P1 + P2 + P3 . . . + Pn
total pressure = 102.6 kPa+ 1.94 kPa
                        = 104.54 kPa
  n = PV / RT
     = (104.54 kPa) (O.151 L) / (0.0821)(290)
     = 15.785 / 23.809
     = 0.663 mols / 60 s
     = 0.0111 mol per second.

[AWK]

Write down  correct balanced equation(s) (temperature is sufficiently low only O2 and traces of water are gaseous).

p(H2O) + p(O2) = 102.6 kPa

[Marcus Soutlo]

Balanced equation is NaI + 2H2O2 --> NaI + 2H2O + O2

p(H2O) + p(O2) = 102.6 kPa --------------------please tell me why you did not add the partial pressure of water vapour- does not the question tell you to account for the partial pressure of water vapour.

[AWK]

Equation is wrong. There are two competetive reaction that strongly depend on pH of solution.
H2O2 + KI = H2O + KOH + I2 without oxygen liberation (need acidic conditions to proceed smothly) and decomposition of H2O2 probably catalysed by traces of iodine
H2O2 = H2O + O2
(both reaction are unbalanced)
But, frankly saing,  you do not need any equation at all. This is a simple problem concerning a gaseous mixture.
From your data is not clear if 151 mL is a pure oxygen or oxygen and a water vapor.
As I understand your problem, the oxygen was collected over water and its volume contains also the water vapor.
From your data (151 mL, 102.6 kPa and T=290 K) calculate moles of both gases then split them proportionally to partial pressures (H2O = 1.94 kPa and O2 = 102.6 - 1.94 kPa.
Finaly divide moles of O2 by 60.

[Borek]

Balanced equation is NaI + 2H2O2 --> NaI + 2H2O + O2

Apart from AWK comment - NaI is on both sides. Cancel it.

[Marcus Soutlo]

Here is additional information I am given about the lab

Purpose:  The purpose of this experiment is to:
a)   determine the order of the reaction with respect to hydrogen peroxide and iodide ion.
b)   determine the rate law equation and the rate constant for the  decomposition of hydrogen peroxide catalyzed by iodide ion.


You have collected oxygen gas by the downward displacement of water (results are in the last column).  This oxygen was produced in the first 60 s of the reaction.

Does this help?

[Marcus Soutlo]

This is what I have done please, by all means, tell me if anything is incorrect.

PV = nRT

n = PV / RT

= (102.6 kPa - 1.94 kPa) (0.151 L) / (0.0821)(290 K)

= (100.6 kPa)(0.151 L) / 23.809

= 15.19 / 23. 809

= 0.6379 mol O2 / min

= 0.6379 mol O2 / 60

= 0.0106 mol O2 / second








PV = nRT

n = PV / RT

= (1.94 kPa) (0.151 L) / (0.0821)(290 K)

=  0.2929 / 23.809

=0.0123 mol H2O / minute

[AWK]

Use correct R constant

[Marcus Soutlo]

PV = nRT

n = PV / RT

= (102.6 kPa - 1.94 kPa)(0.151 L) / (8.314)(290 K)

= (100.6 kPa)(0.151 L) / 2411.205

= 15.19 / 2411. 205

= 0.006299 mol / min

= 0.006299 mol / 60

= 0.00010500 mol / second



Is it just me, or does using 0.0821 for the R value make more sense than using 8.314 for the R value?

[sjb]

Is it just me, or does using 0.0821 for the R value make more sense than using 8.314 for the R value?

It's not simply a matter of making sense (although that of course is useful on occasion). The values of the figures you have given are in kPa, so the value and the units of the gas constant, R, must also match that. I could just as easily say 0.7302 is a better value, or 286.9 - given appropriate units they're fine. Just not in this case.

See http://en.wikipedia.org/w/index.php?title=Gas_constant&oldid=197695421 for values of R in different units.

S

[Marcus Soutlo]

Perhaps your right sjb...but.... 0.006299 mol / min...isn't that a little slow...

[sjb]

Perhaps your right sjb...but.... 0.006299 mol / min...isn't that a little slow

Not especially slow, no - I have done reactions that take days on the mmol scale, as trying to increase the rate by heating etc. caused over reaction.

S