For the decompostion of gaseous dinitrogen pentoxide, 2N2O5(g) ===> 4NO2(g) +O2(g), the rate constant is k=2.8x10^-3 s^-1 at 60 degrees C. The initial concentration of N2O5 is 1.58 mol/L.  ...

1 Answer | Add Yours

jerichorayel's profile pic

Posted on

`2 N_2O_5_(g) -> 4 NO_2_(g) +O_2_(g)`

The rate constant has a unit of `s^-1` which signifies that the reaction order is in the first order. In order to make the reaction equation to first order, let us simplify the coefficients of the species involved.

`2 N_2O_5_(g) -> 4 NO_2_(g) +O_2_(g)`

`N_2O_5_(g) -> 2 NO_2_(g) +1/2 O_2_(g)`

`rate = k[N_2O_5]`

Next, we have to get the rate of the reaction using the rate law (rate equation)


`rate = k[N_2O_5]`

`rate =2.8x10^-3 sec^-1 *1.58 (mol)/(L)`

`rate = 4.424x10^-3 (mol)/(L-sec)`


To get the concentration after 5 minutes, 

`[N_2O_5] = 4.424x10^-3 (mol)/(L-sec) * 5 min * (60 sec)/(1 min)`

`[N_2O_5] = 1.3272 (mol)/(L)`



We’ve answered 323,822 questions. We can answer yours, too.

Ask a question