## Homework question 15.17

$\frac{d[R]}{dt}=-k; [R]=-kt + [R]_{0}; t_{\frac{1}{2}}=\frac{[R]_{0}}{2k}$

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Emily_Bennett_3C
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### Homework question 15.17

When solving for each reactant and the overall order of the reaction, how do we know which experiments to use? And once you set up the fraction ex. (20^a)/10, how do you get 2^a=2?
Thank you.

Chem_Mod
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### Re: Homework question 15.17

You want to choose 2 experiments where only one of the concentrations differs from the other. In this case, they chose 1 and 4 because the concentrations of A and B are the same, but C differs. Then you can find the rate with respect to C. The next 2 experiments you want to choose should vary with either A or B but not both. In this case, they chose experiments 2 and 4 where [A] varies but [B] is the same for both.

The solutions manual has a typo. The exponent a should apply to the entire fraction 20/10 (same with the (200/100)b ). So, your equation should be (20/10)a = 4/2, and then you simplify to get 2a = 2

Michelle Steinberg2J
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### Re: Homework question 15.17

For this question, I tried solving for the order of [C] and got stuck. The solutions manual says that "C is independent of the rate." What does this mean?

mhuang 1E
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### Re: Homework question 15.17

Michelle Steinberg2J wrote:For this question, I tried solving for the order of [C] and got stuck. The solutions manual says that "C is independent of the rate." What does this mean?

This means that C is a zero order reaction: it doesn't depend (or is affected) by the rate. When you compare experiments 1 and 4, you can see that although the initial concentrations of A and B are the same, the initial concentration of C is changed. However, the initial rate for both of the rxns are the same. This shows that "C is independent of the rate."

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