## 15.16

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

Katelyn 2E
Posts: 35
Joined: Sat Jul 22, 2017 3:01 am

### 15.16

This isn't a homework problem, but could someone help me understand this question please? Thank you!

In the reaction 4 Fe2+(aq) + O2(g) + 4 H3O+(aq) --> 4 Fe3+(aq) + 6 H2O(l), when the Fe2+ concentration alone was
doubled, the rate increased by a factor of 8; when both the Fe2+ and the O2 concentrations were increased by a factor of 2, the
rate increased by a factor of 16. When the concentrations of all three reactants were doubled, the rate increased by a factor of
32. What is the rate law for the reaction?

John Huang 1G
Posts: 46
Joined: Thu Jul 13, 2017 3:00 am
Been upvoted: 1 time

### Re: 15.16

I am not too sure on how to make a table chart in chemistry community, but for this problem, it helps to make a table with hypothetical experiments, initial concentrations and initial rates.
This is my best try for a table:
experiment # | [Fe2+] | [O2] | [H30+] | Initial Rate |
1 | 1 | 1 | 1 | 1 |
2 | 2 | 1 | 1 | 8 |
3 | 2 | 2 | 1 | 16 |
4 | 2 | 2 | 2 | 32 |

using the first clue, you can infer that the rate order of Fe2+ is 3 because when the concentration doubles, the rate is (2)^3, which is 8.

To find the rate order of O2, compare experiments 2 and 3. The initial rate doubles when the concentration of O2 doubles, so the rate order of O2 is 1.

To find the rate order of H3O+, compare experiments 3 and 4. The initial rate doubles when the concentration of H3O+ doubles, so the rate order of H3O+ is 1.

Therefore, the rate law would be: RATE: k[Fe2+]^3 [O2] [H3O+].