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Topic: Rate Law  (Read 2538 times)

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Offline futurebrewer

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Rate Law
« on: August 12, 2012, 04:34:57 PM »
Hello, I am doing a lab tomorrow called "Rate Law for the Iodine Clock Reaction." I am supposed to turn the introduction questions tomorrow, but I cannot figure this out.

Consider the following reaction which is used to make millions of tons of ammonia each year.

N2 (g) + 3H2 (g) ------> 2NH3 (g)


There are two questions for this part.

1.) What is the change in concentration of hydrogen per s?

2.) What is the change in concentration of ammonia per s?

I am pretty stumped. I'm thinking maybe negative delta [H2]/s

I honestly have no clue. Any help would be appreciated.

Offline JustinCh3m

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Re: Rate Law
« Reply #1 on: August 12, 2012, 09:44:45 PM »
well, at time just instantly after t=0, if negative delta [H2]/s is the rate of consumption of the reactant H2, using the mole ratio, it follows that the rate of PRODUCTION of NH3 per second would be...

No calculations are needed here

Offline futurebrewer

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Re: Rate Law
« Reply #2 on: August 12, 2012, 11:38:59 PM »
negative delta [NH3]/s?

Offline futurebrewer

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Re: Rate Law
« Reply #3 on: August 13, 2012, 12:51:15 AM »
Ok. I really need some more help. First off, do students usually find Kinetics stuff difficult. I am having a hard time with this. Also, I am quite a bit older. I graduated high school a long time ago, so it's been awhile.

Hydrogen sulfide is a poisonous and bad-smelling pollutant.  One way to remove it from water is to treat the water with chlorine.

H2S (aq) + Cl2 (aq) --------> S (s) + 2H^+ (aq) + 2Cl^- (aq)

Using the following experimental observations, write a rate law for this reaction.

a.) Doubling the concentration of H2S while keeping the concentration of Cl2 constant results in the initial rate increasing by a factor of two.

b.) Halving the concentration of Cl2 while keeping the concentration of H2S fixed results in the initial rate decreasing by a factor of two.

Thanks again for all the *delete me*

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