Question

For the reaction, `2N_(2)O_(5)to4NO_(2)+O_(2)` rate and rate constant are `1.02xx10^(-4) M sec^(-1)` and `3.4xx10^(-5)sec^(-1)` respectively, the concentration of `N_(2)O_(5)`, at that time will be
A. `1.732`
B. `0`
C. `1.02xx10^(-4)`
D. `3.4xx10^(5)`

Answer 1

Correct Answer - B
`2N_(2)O_(5) to 4NO_(2)+O_(2)`
`(-d[N_(2)O_(5)])/(dt)=K.[N_(2)O_(5)]`
`1.02xx10^(-4)=3.4xx10^(-5)A^(-1)xx[N_(2)O_(5)]`
`therefore [N_(2)O_(5)]=(1.02xx10^(-4))/(3.4xx10^(-5))=3`