Berkeley Review Kinetics Question

[0Complications]

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Ok, so I'm doing the problems in the BR Kinetics section 9 of the chemistry books and I'm confused as to how they are deriving their answer (page 204).


The question asks:

Given that S2O3 (aq) in a 0.50 liter flask is consumed at the rate of 0.0080 moles per second, what is the formation rate of S4O6 (aq)?

2 S2O3(aq) + I2(aq) ------> S4O6(aq) + 2 I(aq)

A) 0.0080 M/s
B) 0.0160 M/s
C) 0.0320 M/s
D) 0.0160 s/M


So, I've selected the wrong answer. Does anyone know what the correct answer is and why? I have the correct answer as listed in the book, but I think I'll get more beneficial responses if I wait to post the correct answer. This way I won't get the, "Duh dude, it is this..." Maybe I will anyway, but any help is appreciated.

 

[capn jazz]

So S2O3 is consumed at .008 moles/.5 L/s. You need the actual molar relationship, so that would be .016M/s. Since for every 2 S2O3 molecules consumed, 1 S4O6 molecule is produced, you divide .016M/s by 2 to get .008 M/s.

 

[Dr Gerrard]

0.008 mol/s S2O3 * (1 mol S4O6/2 mol S2O3) = 0.004 mol/s of S4O6

You want concentration.

If you get 0.004 additional mol/s of S4O6, that would mean that the concentration changes by this divided the total volume.

THus, (0.004 mol/s)/ 0.5 L = 0.008 M/s

So basically, what Capn Jazz said.

 

[0Complications]

Thanks guys.


I think the problem I'm having with this question is that the stoichiometric relationships between the reactant side and the product side are 2:1, so I was thinking that in order to produce 1 mole of product you would need 2 moles of the reactant. This is what led me to believe the answer was B (0.0160 M/s), otherwise I don't see how you would have enough reactant to form the product in half the time that is needed to for the reactants to shift to products.

I don't know if that makes sense, but apparently it shouldn't... I'll play around with it a bit or just rely on the math, which I really don't like doing.

 

[0Complications]

0.008 mol/s S2O3 * (1 mol S4O6/2 mol S2O3) = 0.004 mol/s of S4O6

You want concentration.

If you get 0.004 additional mol/s of S4O6, that would mean that the concentration changes by this divided the total volume.

THus, (0.004 mol/s)/ 0.5 L = 0.008 M/s

So basically, what Capn Jazz said.

NM. Strike what I just said. This makes sense. The bolded part is the step I missed/didn't understand. Thanks!