Question

13) For the reaction: 2N2O5(g) ?4NO2(g) +O2(g) the rate law is: ?[O2] ?t = k[N2O5] At 300 K, the half-life is 2.50

Answer 1

answer A, t2 = 6.57x10^3 s, is correct.

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first of all.. activation energy and temperaure suggests the Arrhenius equation..
k = A exp (-Ea/RT)

and you have rate law so you we're going to have to combine the two.. and this will involve an integration...ok?

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from this equation...
2N2O5(g) -> 4NO2(g) + O2(g)

you can see that for every 2 N2O5 you get 1 O2...meaning...
-1/2 x d[N2O5] = d[O2]... right?... 1/2 x change in N2O5 = change in O2?
and...N2O5 is decreasing while O2 is increasing? hence the "-" sign...

ie...
d[O2] / dt = k[N2O5]

can be written as...
-1/2 x d[N2O5] / dt = k[N2O5]

rearranging...
(1 / [N2O5]) d[N2O5] = - 2 x k x dt
right?

maybe it would help to let A = [N2O5]?
(1/A) dA = -2 k dt
??.. is that easier to see?

anyway... integrating...
?(1 / [N2O5]) d[N2O5] = ?(- 2 x k x dt)

gives this...
ln[N2O5]@t - ln[N2O5]initial = -2kt

rearranging...via log rules...
ln( [N2O5]@t / ln[N2O5]initial) = -2kt

and half life means this...[N2O5]@t = 1/2 [N2O5] initial...
ie...the concentration of N2O5 has dropped in half...

so that...
ln( [N2O5]@t / ln[N2O5]initial) = -2kt
becomes...
ln(1/2 [N2O5]initial / [N2O5]initial) = -2kt
ie...
ln(1/2) = -2kt
ie...
-ln(2) = -2kt
ie...
ln(2) = 2kt

ok.... now what?...
well...
we know that...
k = A exp (-Ea/RT)

so....by substituting...
k = A exp (-Ea/RT)
into...
ln(2) = 2kt
we get...
ln(2) = 2 x (A exp (-Ea/RT)) x t

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now you can solve for A at T = 300K and then recalculate t at T=310K if you like but... I'll take a slightly different approach...

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what if we rearranged the equation like this...
ln(2) / (2 exp (-Ea/RT)) x t) = A

and we assumed A and Ea are constant... so that...
ln(2) / (2 exp (-Ea/RT1)) x t1) = A = ln(2) / (2 exp (-Ea/RT2)) x t2)

rearranging...and canceling...
exp (-Ea/RT2)) x t2 = exp (-Ea/RT1)) x t1
t2 = t1 x exp(-Ea/RT1) / exp(-Ea/RT2)
t2 = t1 x exp(-Ea/RT1 - (-Ea/RT2))
t2 = t1 x exp(Ea/R x (1/T2 - 1/T1))

and that's probably about as simplfied as I'm going to get so...
t2 = (2.50x10^4s) x exp((103300kJ/mole / 8.314 J/moleK) x (1/310K - 1/300K))
t2 = (2.50x10^4s) x .2629
t2 = 6.57x10^3 s

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questions?

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notice Kirby's rate equation "ln (2) /t(h) = k" is this...
ln(2) = kt

where mine is this..
ln(2) = 2kt

Kirbys error was because you can't integrate.
d[O2]/dt = k[N2O5]
directly...
you have to convert the equation so the units match up. ie.. [O2] ---> [N2O5]
and d[O2]/dt ? d[N2O5]/dt

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one other thing... does this remind you of anything?
t2 = t1 x exp(Ea/R x (1/T2 - 1/T1))
????

what if I did this...
t2/t1 = exp(Ea/R x (1/T2 - 1/T1))
then.. natural logging...
ln(t2/t1) = (Ea/R) x (1/T2 - 1/T1))

does that look anything like...
ln(P1/P2) = (Ea/R) x (1/T2 - 1/T1))
?