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Rates and Orders Question.
Can someone help me understand this.
So the question follows :
One cause of low-level smog is the reaction of ozone, O3, with ethene, C2H4. The smog contains methanal, HCHO(g).
The equation for methanal production is shown below.
O3(g) + C2H4(g) ----> 2HCHO(g) + ½ O2(g)
The rate of the reaction was investigated, using a series of different concentrations of either C2H4(g) or O3(g), by measuring the initial rate of formation of HCHO(g).
The results are shown below.
In part i) I got the rate expression rate = k[O3(g)][C2H4(g)]
In part ii) I was asked to work out the rate constant, what I got is 2.0x10^-12 dm3 mol-1 s-1.
I don't understand part iii) which is
Using the equation above, deduce the initial rate of formation of O2(g) in experiment 1.
Explain your reasoning.
Can someone explain how to work this out, I've looked in the mark scheme nothing makes sense.
So the question follows :
One cause of low-level smog is the reaction of ozone, O3, with ethene, C2H4. The smog contains methanal, HCHO(g).
The equation for methanal production is shown below.
O3(g) + C2H4(g) ----> 2HCHO(g) + ½ O2(g)
The rate of the reaction was investigated, using a series of different concentrations of either C2H4(g) or O3(g), by measuring the initial rate of formation of HCHO(g).
The results are shown below.
In part i) I got the rate expression rate = k[O3(g)][C2H4(g)]
In part ii) I was asked to work out the rate constant, what I got is 2.0x10^-12 dm3 mol-1 s-1.
I don't understand part iii) which is
Using the equation above, deduce the initial rate of formation of O2(g) in experiment 1.
Explain your reasoning.
Can someone explain how to work this out, I've looked in the mark scheme nothing makes sense.
Reply 1
its just the wording...isnt it just rate=k x (0.5 x 1)
Reply 2
What does the mark scheme say?
Reply 3
you know from the equation that o2 is being formed, so you use the formula u came up with ( rate = k[O3(g)][C2H4(g)] ) to work it out.
Reply 6
Firstly, your rate constant isnt correct, notice the powers of ten.
Rate of reaction is change in concentration of product per unit time. Now look at the moles in your equation, 4 moles of methanal are produced for every 1 mole of oxygen.
Rate=Delta(concentration) / time
Time is constant but concentration of oxygen is 0.25 that of methanal. So initial rate for methanal/4 gives initial rate for O2(g).
Rate of reaction is change in concentration of product per unit time. Now look at the moles in your equation, 4 moles of methanal are produced for every 1 mole of oxygen.
Rate=Delta(concentration) / time
Time is constant but concentration of oxygen is 0.25 that of methanal. So initial rate for methanal/4 gives initial rate for O2(g).
Reply 7
uer23
mark scheme answer is 2.5 x 10 ^ -13
Apparently they divided the initial rate in experiment 1 (1.0 x 10 ^ -12) by 4. I don't get why
Apparently they divided the initial rate in experiment 1 (1.0 x 10 ^ -12) by 4. I don't get why
well theyre giving the rate in terms of the HCHO but they want it in terms of O2 and for every 2 moles of HCHO produced you have 1/2 mole of O2 so you just divide it by 4 to get the rate for the formation of O2 instead of HCHO
Reply 8
how did you even work out the rate equation? - the mark scheme doesn't make any sense to me=[
(edited 14 years ago)
Original post by uer23
Can someone help me understand this.
So the question follows :
One cause of low-level smog is the reaction of ozone, O3, with ethene, C2H4. The smog contains methanal, HCHO(g).
The equation for methanal production is shown below.
O3(g) + C2H4(g) ----> 2HCHO(g) + ½ O2(g)
The rate of the reaction was investigated, using a series of different concentrations of either C2H4(g) or O3(g), by measuring the initial rate of formation of HCHO(g).
The results are shown below.
In part i) I got the rate expression rate = k[O3(g)][C2H4(g)]
In part ii) I was asked to work out the rate constant, what I got is 2.0x10^-12 dm3 mol-1 s-1.
I don't understand part iii) which is
Using the equation above, deduce the initial rate of formation of O2(g) in experiment 1.
Explain your reasoning.
Can someone explain how to work this out, I've looked in the mark scheme nothing makes sense.
So the question follows :
One cause of low-level smog is the reaction of ozone, O3, with ethene, C2H4. The smog contains methanal, HCHO(g).
The equation for methanal production is shown below.
O3(g) + C2H4(g) ----> 2HCHO(g) + ½ O2(g)
The rate of the reaction was investigated, using a series of different concentrations of either C2H4(g) or O3(g), by measuring the initial rate of formation of HCHO(g).
The results are shown below.
In part i) I got the rate expression rate = k[O3(g)][C2H4(g)]
In part ii) I was asked to work out the rate constant, what I got is 2.0x10^-12 dm3 mol-1 s-1.
I don't understand part iii) which is
Using the equation above, deduce the initial rate of formation of O2(g) in experiment 1.
Explain your reasoning.
Can someone explain how to work this out, I've looked in the mark scheme nothing makes sense.
hello there, I am a student taking this exam now and have been doing the same question, I find it impossible to understand, I mean could you please help me with why they divided it by 4? Thank you
so the equation is:
O3 + C2H4 = 2HCHO + 0.5O2
as the question is asking for O2 not 0.5O2 you times the whole equation by 2
so youll get 4HCHO + O2
in experiment 1 the the initial rate of HCHO is 1X10 power -12
so u divide that by 4 to get the initial rate of O2
so u get 2.5 X 10 power -13
O3 + C2H4 = 2HCHO + 0.5O2
as the question is asking for O2 not 0.5O2 you times the whole equation by 2
so youll get 4HCHO + O2
in experiment 1 the the initial rate of HCHO is 1X10 power -12
so u divide that by 4 to get the initial rate of O2
so u get 2.5 X 10 power -13
Reply 11
This makes sense.Thanks
Reply 12
Original post by suzaie
so the equation is:
O3 + C2H4 = 2HCHO + 0.5O2
as the question is asking for O2 not 0.5O2 you times the whole equation by 2
so youll get 4HCHO + O2
in experiment 1 the the initial rate of HCHO is 1X10 power -12
so u divide that by 4 to get the initial rate of O2
so u get 2.5 X 10 power -13
O3 + C2H4 = 2HCHO + 0.5O2
as the question is asking for O2 not 0.5O2 you times the whole equation by 2
so youll get 4HCHO + O2
in experiment 1 the the initial rate of HCHO is 1X10 power -12
so u divide that by 4 to get the initial rate of O2
so u get 2.5 X 10 power -13
hi i wanted to ask why do you divide by 4?
Reply 13
Original post by sumzie2003
hi i wanted to ask why do you divide by 4?
the ratio of HCHO : O2 is 4:1
so if the initial rate of 4HCHO is 1x10^-12
then the initial rate of O2 will be 1x10^-12 / 4
Reply 14
Hi I know im 8 years too late but can someone help me understand this? I can see how this works but I feel like if I saw this in an exam I would try to use [O3(g)], rather than [C2H4(g)], which wouldn't work as you would divide the same rate in experiment 1 by 2, rather than 4. I was just wondering If someone could explain to me how I would know to use [C2H4]. Thanks!
Original post by suzaie
so the equation is:
O3 + C2H4 = 2HCHO + 0.5O2
as the question is asking for O2 not 0.5O2 you times the whole equation by 2
so youll get 4HCHO + O2
in experiment 1 the the initial rate of HCHO is 1X10 power -12
so u divide that by 4 to get the initial rate of O2
so u get 2.5 X 10 power -13
O3 + C2H4 = 2HCHO + 0.5O2
as the question is asking for O2 not 0.5O2 you times the whole equation by 2
so youll get 4HCHO + O2
in experiment 1 the the initial rate of HCHO is 1X10 power -12
so u divide that by 4 to get the initial rate of O2
so u get 2.5 X 10 power -13
(edited 2 years ago)
Reply 15
This one stumped me for a bit too, but I think I get it now? If you look in the main body question it says "measuring the initial rate of formation of HCHO(g)." AKA, the initial rates in the table are all for the formation of HCHO. In order to get the rate of O2, you need to divide the rate of HCHO by how many more moles it has than O2. The products are 2HCHO and 1/2O2, so essentially a 4:1 ratio, hence you divide the rate by 4.
Original post by juliestudies
Hi I know im 8 years too late but can someone help me understand this? I can see how this works but I feel like if I saw this in an exam I would try to use [O3(g)], rather than [C2H4(g)], which wouldn't work as you would divide the same rate in experiment 1 by 2, rather than 4. I was just wondering If someone could explain to me how I would know to use [C2H4]. Thanks!
Reply 16
This one stumped me for a bit too, but I think I get it now? If you look in the main body question it says
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