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For the decompostion of gaseous dinitrogen pentoxide, 2N2O5(g) ===> 4NO2(g) +O2(g),...

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kathyyyyyy123 | Student, Undergraduate | Honors

Posted July 12, 2013 at 4:56 AM via web

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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.

 

What is [N2O5] after 5.00 mins?

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jerichorayel | College Teacher | (Level 1) Senior Educator

Posted July 12, 2013 at 5:58 AM (Answer #1)

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`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)`

 

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