# Rates of chemical reactions I : A clock reaction

S2O8 + 2I >>> I2 + 2SO4

what effect does doubling the concentration of iodine have on the rate of this reaction ?

what effect does changing the [S2O8] have on the reaction ?

I tried to calculate the x and y but I get negative answers and this impossible , where will be the wrong in my calculation ?

 Asoom Posts: 52, Reputation: 1 Junior Member #2 Apr 20, 2009, 11:53 AM

Nobody want help me?!
 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #3 Apr 20, 2009, 11:57 AM

Be a little patient...

Ok, first thing, you forgot one product in your equation, solid sulfur.

What have you learnt? Increased surface area increases the rate of reaction. Is bubbling increasing the surface area?

The second question, I don't quite understand... Post back your question in a clearer way please.

I don't see x or y in your question?

Ok, will be here later on... bedtime for me now.

EDIT : I made a mistake there.. there are no sulfur
 Asoom Posts: 52, Reputation: 1 Junior Member #4 Apr 20, 2009, 01:07 PM

Thanks for your helping and have a nice dream

I mean in the second one that if we change the concentration of S2O8 how this change will effect in the reaction..

The third Q: I get the rate from the graph

1- 4*10^-4
2- 3*10^-4
3- 2*10^-4
4- 5*10^-4

a- [S2O8] = 0.05 , [i] = 0.05
b- [S2O8] = 0.10, [i] = 0.05
c- [S2O8] = 0.05 , [i] = 0.10
d- [S2O8] = 0.10, [i] = 0.025

so when I calculate the x ,y I get negative answer

x=-0.41
y=-1
 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #5 Apr 20, 2009, 08:12 PM

3. Ok, what have you learnt when you increase the concentration of a reactant?

Increasing the concentration is increasing the number of molecules of S2O8. Therefore, the collisional frequency is increased. Let's say there is a fine rain hitting the ground. The 'concentration' is the amount of raindrops hitting a certain area. If the rain becomes a downpour, there are more drops hitting the same area, and the ground gets wet faster! So, increasing concentration will increase the rate of reaction.

4. I think I did something wrong yesterday... was too tired I think. Let me re write the equation (you are sure this is right?)

$S_2O_8 +I_2 --\rightarrow 2I + 2SO_4$

The 4th question is still unclear. Was there x in front of S2O8 and y in front of I2 in the equation? If so, you need to consult your graph. Find the point where you obtained the maximum rate of reaction. This point should give your values of x and y.
 Asoom Posts: 52, Reputation: 1 Junior Member #6 Apr 20, 2009, 11:15 PM

Thanks a lot , its clear now

yes I think the equation was right

 Asoom Posts: 52, Reputation: 1 Junior Member #7 Apr 20, 2009, 11:16 PM

Thanks a lot , its clear now

yes I think the equation was right

 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #8 Apr 21, 2009, 04:37 AM

Ha! You double posted... try not to do that again. Glad that you got all the answers! Cya!

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