## 7A 15

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

Posts: 55
Joined: Fri Sep 28, 2018 12:26 am

### 7A 15

How do you know that molecule C is a zero order reaction? Also how do you calculate the degree for each reaction?

Elizabeth Gallmeister 1A
Posts: 41
Joined: Fri Sep 28, 2018 12:21 am

### Re: 7A 15

If you look at experiment 1 and 4, the only concentration that changes is for C. Yet, if you compare the initial rate for both experiments, they're the same value (2.0). So, we can see that the change in concentration of C has no bearing on the rate of the reaction, and therefore it is raised to the 0th power.

tierra parker 1J
Posts: 61
Joined: Fri Sep 28, 2018 12:17 am

### Re: 7A 15

to calculate the degree for each reactant you have to compare each experiment and see how changing the concentration of one reactant impacts the total rate of the reactant. for element c, you can see in experiment 1 and experiment 4 the concentration of a and b aren't change but the concentration of c went from 700 to 400 but the rate of the reaction stayed at 2.0. this means that the concentration of c has no effect on the total rate which makes a zero order reaction

vaishali 1D
Posts: 61
Joined: Fri Sep 28, 2018 12:19 am

### Re: 7A 15

By looking at the first and fourth experiment, you can see that the only concentration change is C but the initial rate in both those experiments are the same, therefore concentration doesn't change the rate of the reaction and that's the definition of a zero order reaction.

Jessica Chen 1F
Posts: 61
Joined: Fri Sep 28, 2018 12:17 am

### Re: 7A 15

You can tell that molecule C is a zero order reaction because when you divide the equations (you select the 2 equations where the concentration of A and B are the same and only C is different, which would be in experiment 1 and 4), you get 2/2 = (7/4)^z

Since 2/2 is 1, Z has to be 0 in order for (7/4)^z to equal 1.