[skyelite00]

I am not sure how to answer this question.

Q: For the reaction 2H2O2 ---> 2H2O + O2, that is catalysed by an excess of Fe^3+, calculate the maximum amount of O2 gas produced (in mL) at ambient temperature (T = 300 K) from 30ml of H2O2 solution at 0.2 mol L-1.

[sjb]

I'm not sure this is kinetics, more stoichiometry. How much hydrogen peroxide do you have? How much oxygen can that produce from the balanced equation?

[skyelite00]

I'm not sure this is kinetics, more stoichiometry. How much hydrogen peroxide do you have? How much oxygen can that produce from the balanced equation?

Well there is 30ml of H202 solution at 0.2 mol L-1.
Using n = c/v
         n = 0.2 mol L-1  /  0.030 L
         n = 6.67 moles of H2O2

The number of moles of O2 should be half of H2O2.
As 2H202 ---> 2H20 + O2 therefore O2 = 6.67 mol / 2
O2 = 3.33 moles of Oxygen produced.

To find the volume for Oxygen we can use the equation: n1v1 = n2v2
(6.67 mol * 0.030 L)  / (3.33 * V2)
V2 = 0.2001 / 3.33
V2 = 0.060 L
V2 = 60 ml

Therefore the oxygen produced is 60 ml.

Is this correct? I just realised i typed out the wrong question.

[chenbeier]

I would think again how much moles of H2O2 you have. Your answer  n=c/V is wrong. Check the units. In your case it would be mol/l². The second thing is you forgot the temperature in your calculation.

[skyelite00]

Sorry for the confusion the actual question is the following:

a) The reaction between hydrogen and nitric oxide ( at 700 C) is given by the following expression:
2H2 + 2NO ---> 2H20 + N2

If NO is observed to disappear at a rate of 2.1 x 10-6 Ms-1,
(i) What us the rate of appearance of N2? Justify your answer.
(ii) What is the overall rate of reaction? Justify your answer.


b) For the 2N2O5 ---> 4NO2 + O2 decomposition reaction the rate constant (k) was measured at five different temperature. The plot of ln k vs 1/T nicely fits with the slope m = -1029 K

(i) What is the value of Ea for this reaction kJ/mol?

[skyelite00]

I would think again how much moles of H2O2 you have. Your answer  n=c/V is wrong. The second thing is you forgot the temperature in your calculation.

Actually wait i think i may have got it.

[skyelite00]

moles H2O2 = 0.0030L * 0.2 mol.L-1
=0.0006 mol H202
= 0.0003 mol O2

V= nRT/P
V= (0.0003 mol O2 * 0.00820578 L.atm.K-1.mol-1 x 300K ) / 1 atm

= 0.0007385 L O2
= 7.38 ml O2

Do i need to calculate excess [Fe3+]?

[Borek]

Do i need to calculate excess [Fe3+]?

It is a catalyst, its amount doesn't matter for the stoichiometry.