## 15.17

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

Katherine Jordak 1H
Posts: 51
Joined: Fri Sep 29, 2017 7:05 am

### 15.17

I'm confused about 15.17 part a. In the answer guide it says use experiments 1 and 4 to show that [C] is independent of the rate. Can someone explain to me how I can deduce [C] is independent of the rate?

Andy Liao 1B
Posts: 52
Joined: Thu Jul 13, 2017 3:00 am

### Re: 15.17

We compare Experiments 1 and 4 to find the order with respect to C because [A] and [B] are the same concentration in both (10. mmol.L^-1 and 100. mmol.L^-1, respectively). We know that [C] is independent of the rate because even though the concentration of C in Experiment 4 is different from the one in Experiment 1, the initial rate is unchanged. The initial rate is 2.0 mmolG.L^-1.s^-1 in both.

104922499 1F
Posts: 53
Joined: Fri Sep 29, 2017 7:04 am

### Re: 15.17

you use experiments 1 and 4 to find [C] in respect to [A] and [B] because A and B values are the same values. This helps in finding the order of [C] because the values for [C] in experiment 1 and 4 are different.

Gurkriti Ahluwalia 1K
Posts: 37
Joined: Fri Sep 29, 2017 7:07 am

### Re: 15.17

therefore, is it safe to say that you DO NOT calculate the order of C because it is independent of the rate?? in other words, only the orders of A and B matter because they affect the rate

Matthew Lin 2C
Posts: 30
Joined: Fri Sep 29, 2017 7:06 am

### Re: 15.17

Gurkriti Ahluwalia 1K wrote:therefore, is it safe to say that you DO NOT calculate the order of C because it is independent of the rate?? in other words, only the orders of A and B matter because they affect the rate

Well when you determine that the concentration of C is independent of the reaction rate, then you are essentially calculating the order of C since you know therefore that its order is zero. After you figure this out, then you can ignore C and just focus on the concentrations of A and B.