Answer:
4600s
Explanation:
[tex] 2N_{2}O_{5}(g) - - -> 4NO_{2}(g) +O_{2}[/tex]
For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:
[tex]-\frac{d[B]}{dt}=k[B] - - - -\frac{d[B]}{[B]}=k*dt[/tex]
If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.
PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.
[tex]-\frac{d[P(B)]}{P(B)}=k*dt[/tex]
[tex]-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt[/tex]
Integrating we get:
[tex]\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt[/tex]
[tex]-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})[/tex]
Clearing for t2:
[tex]\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}[/tex]
[tex]ln[P(N_{2}O_{5})]=ln(650)=6.4769[/tex]
[tex]ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333[/tex]
[tex]t_{2}=\frac{-(6.4769-6.6333)}{3.4*10^{-5}}+0= 4598.414s[/tex]
