x Turn on thread page Beta
 You are Here: Home

# Kinetics question watch

1. This question is quite easy save for the last part, but I need to show the earlier parts otherwise you won't be able to help.

The Following reaction
2N2O5 ---> 4NO2 + O2

is described as first order with respect to N2O5

a) Write down the rate expression for the reaction

b) Explain what is meant by the term half life for this reaction
answer: Time taken for the concentration of N2O5 to fall to half its initial value

c) State what is characteristic about the half life of a first order reaction
answer: the half life is constant

d) The rate constant at a certain temperature for the reaction is 0.0052 (standard units) calculate the time taken for the concentration of N2O5 to decrease from 0.1 mol dm-3 to 0.01 mol dm-3

answer: ?? (my intelligent guess is 5.2 seconds..)
I would be very grateful for any help!
2. (Original post by andymt)
This question is quite easy save for the last part, but I need to show the earlier parts otherwise you won't be able to help.

The Following reaction
2N2O5 ---> 4NO2 + O2

is described as first order with respect to N2O5

a) Write down the rate expression for the reaction

b) Explain what is meant by the term half life for this reaction
answer: Time taken for the concentration of N2O5 to fall to half its initial value

c) State what is characteristic about the half life of a first order reaction
answer: the half life is constant

d) The rate constant at a certain temperature for the reaction is 0.0052 (standard units) calculate the time taken for the concentration of N2O5 to decrease from 0.1 mol dm-3 to 0.01 mol dm-3

answer: ?? (my intelligent guess is 5.2 seconds..)
I would be very grateful for any help!
You have already ascertained that the reaction is first order and that the half life of a first order reaction is constant.

Using the expression t1/2 = 0.693/k

you get t1/2 = 133.3 s

now take it from there...
3. answer: ?? (my intelligent guess is 5.2 seconds..)
I would be very grateful for any help![/QUOTE]

I think as mentioned by charco,

1/100 = (1/2)^n x 1/10

Final conc = no.of half lifes x initial concentration

you would get like n = ln10/ln2

half life = ln2/k(or ln2 =0.693, but better be exact)

so total time taken = no.of half lifes x n
= ln10/k

I think i might have done that correctly, but if someone could tell me what i have done wrong, that'd be appreciated too.
4. so you need to work out how many half lives the decrease in concentration represents? how do you do that?
5. OK the form is ln(N0/Nt) = -kt

Where N0 is the initial amount
and Nt is amount remaining at time t

therefore the time taken is ln(N0/Nt)/k

= ln(0.1/0.01)/0.0052 = 442.8 s
6. (Original post by charco)
OK the form is ln(N0/Nt) = -kt

Where N0 is the initial amount
and Nt is amount remaining at time t

therefore the time taken is ln(N0/Nt)/k

= ln(0.1/0.01)/0.0052 = 442.8 s
Thanks a lot
but why is the minus sign relevant?
7. (Original post by andymt)
Thanks a lot
but why is the minus sign relevant?
You got me thinking and I noticed that I put the relationship upside down!

It should be ln(Nt/N0) = -kt

The way I wrote it before gives you a negative answer (that's why the negative sign is relevant)!

ln(0.01/0.1)/0.0052 = -t

time taken = 442.8 s

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: November 9, 2008
Today on TSR

### University rankings 2019

Cambridge at number one

### I have imposter syndrome at Cambridge

Poll

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE