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  • <img alt="N_{2}O_{5} \rightarrow 2NO_{2} + \frac{1}{2}O_{2}" src="https://learn.careers360.com/latex-image/?N_%7B2%7DO_%7B5%7D%20%5Crightarrow%202NO_%7B2%7D%20&plus;%20%5Cfrac%7B1%7D%7B2%7DO_%7

N_{2}O_{5} \rightarrow 2NO_{2} + \frac{1}{2}O_{2}

When N2O5 decompose, its t_{\frac{1}{2}} does not change with its changing pressure during the reaction. So which one is the correct representation for ''pressure of NO2'' vs ''time'' during the reaction when initial is equals to P0

Option: 1


Option: 2


Option: 3


Option: 4


Answers (1)

best_answer

 

nth order reaction -

The rates of the reaction is proportional to nth power of reactant

- wherein

Differential rate law

=\frac{dx}{dt}=k(a-x)^{n}

Integrated rate laws,

=\frac{1}{n-1}[(a-x)^{1-n} -a^{(1-n)}]=k_nt

a= initial, concentration of reactant at t=0 sec

x= concentration of product formed at t= tsec

t_\frac{1}{2}=\frac{1}{(n-1)(a^{n-1})(k_n)}[2^{n-1}-1]

Formulae for all the order except n=1

 

 

 

Theory Based

Posted by

Riya

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