Question

At 25∘C, the decomposition of dinitrogen pentoxide, N2O5(g), into NO2(g) and O2(g) follows first-order kinetics with k=3.4×10−5 s−1. A sample of N2O5 with an initial pressure of 760 torr decomposes at 25∘C until its partial pressure is 650 torr. How much time (in seconds) has elapsed? At 25, the decomposition of dinitrogen pentoxide, (g), into (g) and (g) follows first-order kinetics with . A sample of with an initial pressure of 760 decomposes at 25 until its partial pressure is 650 . How much time (in seconds) has elapsed? 5.3×10−6 2000 4600 34,000 190,000

Answer 1

4600s

Step-by-step explanation:

`2N_(2)O_(5)(g) - - -> 4NO_(2)(g) +O_(2)`

For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:

`-(d[B])/(dt)=k[B] - - -  -(d[B])/([B])=k*dt`

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.

`-(d[P(B)])/(P(B))=k*dt`

`-(d[P(N_(2)O_(5))])/(P(N_(2)O_(5)))=k*dt`

Integrating we get:

`\int\limits^p \,-(d[P(N_(2)O_(5))])/(P(N_(2)O_(5)))=\int\limits^ t k*dt`

`-(ln[P(N_(2)O_(5))]-ln[P(N_(2)O_(5))_(o))])=k(t_(2)-t_(1))`

Clearing for t2:

`(-(ln[P(N_(2)O_(5))]-ln[P(N_(2)O_(5))_(o))]))/(k)+t_(1)=t_(2)`

`ln[P(N_(2)O_(5))]=ln(650)=6.4769`

`ln[P(N_(2)O_(5))_(o)]=ln(760)=6.6333`

`t_(2)=(-(6.4769-6.6333))/(3.4*10^(-5))+0= 4598.414s`