## Example 15.3

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

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Nha Dang 2I
Posts: 53
Joined: Fri Sep 29, 2017 7:07 am

### Example 15.3

What concentration of N2O5 remains 10.0min (600 s) after the start of its decomposition at 65 C (in reaction A) when its initial concentration was .040 mol/L? See Table 15.1 for the rate law.

Where did they find k which equals 5.2 x 10^-3? Why isn't it 3.7x10^-5?

Jessica Lutz 2E
Posts: 56
Joined: Fri Sep 29, 2017 7:04 am

### Re: Example 15.3

The decomposition is happening at 338K. k changes with temperature, which is why table 15.1 shows different values of k depending on the temperature of the reaction. It would be 3.7x10^-5 if the reaction was happening at 298K.

Shanmitha Arun 1L
Posts: 53
Joined: Fri Sep 29, 2017 7:04 am
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### Re: Example 15.3

Make sure to look at the correct temperature in the table. Generally on tests though, I assume the k will be given but you should still know how to find it based on a table.

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