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Hello

If we have a number of trials, and are given the concentration of 3 reactants is this how you would find the rate law? Also how would you find the average constant?

[tex] 1 : 0.001 \ \ 0.1 \ \ 0.2 \ \ 1.2 \times 10^{-6} [/tex]

[tex] 2 : 0.001 \ \ 0.4 \ \ 0.2 \ \ 0.48\times 10^{-5} [/tex]

[tex] 3 : 0.003 \ \ 0.1 \ \ 0.2 \ \ 108\times 10^{-7} [/tex]

[tex] 4 : 0.003 \ \ 0.4 \ \ 0.4 \ \ 86.4 \times 10^{-6} [/tex]

The 4 numbers are the # of trials, and the three numbers beside them are the concentration of the three reactants. The last number is the rate in [tex] mol/L*s [/tex]

So using the rate law [tex] k[A]^{m}

If we have a number of trials, and are given the concentration of 3 reactants is this how you would find the rate law? Also how would you find the average constant?

[tex] 1 : 0.001 \ \ 0.1 \ \ 0.2 \ \ 1.2 \times 10^{-6} [/tex]

[tex] 2 : 0.001 \ \ 0.4 \ \ 0.2 \ \ 0.48\times 10^{-5} [/tex]

[tex] 3 : 0.003 \ \ 0.1 \ \ 0.2 \ \ 108\times 10^{-7} [/tex]

[tex] 4 : 0.003 \ \ 0.4 \ \ 0.4 \ \ 86.4 \times 10^{-6} [/tex]

The 4 numbers are the # of trials, and the three numbers beside them are the concentration of the three reactants. The last number is the rate in [tex] mol/L*s [/tex]

So using the rate law [tex] k[A]^{m}

**^{n}[C]^{z} [/tex] I know that in trial 1 the rate goes down by a factor of [tex] 1/4 [/tex]. A and C are the same but B changes. Does that mean B's reaction order is 1/4?**

Any help in trying to find the

ThanksAny help in trying to find the

**average**value of the rate constant and the rate law would be appreciated.Thanks

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