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Rates of reactions [math] N_2O_4 \rightarrow 2 NO_2[/math]


Dhamnekar Win,odd

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22 hours ago, Dhamnekar Win,odd said:

image.png.ec0ea46f671a08fa5a42dab579cb2205.png

image.png.eda22cf1039a54eb259040e3a32fafb6.png

How to answer all these questions? 

image.png.0067f814d4f162fed17d81db2ddc1434.png

Are these above answers correct?

Answers to remaining questions will be provided very soon.

 

Agree with your  ( a ) but your ( b ) seems to overlook the fact that the rate has already decayed away to some extent after one minute has elapsed, compared to its initial value, because by this time the concentration of reactant has already fallen appreciably. I think you need to go back to the rate expression for a 1st order reaction: 

- d[A]/dt = k[A], (A being the reactant i.e. consumed with time)

from which you get:

[A]/[A₀] = exp[-kt].

And take it from there.

 

 

Edited by exchemist
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Answer to d)

1)∆H = 17.4 kJ/mol

2)∆H = 8.7 kJ/mol

3)∆H= 4.35 kJ/mol

Answer to e) Mechanism for this reaction is [math]1)N_2 + 2 O_2 \rightarrow 2 NO_2  \Delta H^\circ_f = 66.4 kJ/mol [/math] 

[math]2) N_2O_4 \rightarrow N_2 + 2 O_2  \Delta H^\circ_f = -9.16 kJ/mol [/math]

Edited by Dhamnekar Win,odd
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3 hours ago, Dhamnekar Win,odd said:

Answer to d)

1)∆H = 17.4 kJ/mol

2)∆H = 8.7 kJ/mol

3)∆H= 4.35 kJ/mol

Answer to e) Mechanism for this reaction is 1)N2+2O22NO2ΔHf=66.4kJ/mol  

2)N2O4N2+2O2ΔHf=9.16kJ/mol

Re ( d) I do not believe you can calculate ΔH for the decomposition from the information you have been given. I assume you have to look up some enthalpy values. It is not clear to me what is meant by the 3 different values you quote.

Your ( e ) is a reaction scheme for the formation of N2O4 from oxygen and nitrogen, but it is not a reaction mechanism for the decomposition of N2O4, which I think is what you are being asked for. I think you need a step to activate the molecule and then a step by which it dissociates.

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