Nitrogen pentoxide decomposes according to equation: `2N_(2)O_(5)(g) to 4NO_(2)(g)+O_(2)(g)`
This first order reaction was allowed to proceed at `40^(@)C` and the data below were collected:

(i) Calculate the rate constant. Include units with your answer.
(ii) What will be the concentration of `N_(2)O_(5)` after 100 minutes?
(iii) Calculate the initial rate of reaction?

Explanation

(i) The equation for the first order reaction is
`k=(2.303)/(t)"log"([R]_(0))/([R])`
Substituting the values in the equation above, we have
For t = 20 min, `k=(2.303)/(20"min")"log"(0.4"mol L"^(-1))/(0.289"mol L"^(-1))=(2.303)/(20"min") log 1.3841 =(2.303)/(20"min")xx0.1412`
`=0.01626"min"^(-1)`
For t = 40 min, `k=(2.303)/(40"min")"log"(0.4"mol L"^(-1))/(0.209"mol L"^(-1))=(2.303)/(40"min") log 1.9138 =(2.303)/(40"min")xx0.2818`
`=0.01622"min"^(-1)`
We may take the average of the two values i.e., 0.01624`" min"^(-1)` .
(ii) Again use the relation
`k=(2.303)/(t)"log"([R]_(0))/([R])`
t = 100 minutes, `k=0.01624 " min"^(-1), [R]_(0)=0.400"mol L"^(-1), [R]=`?
Substituting the values in the equation above, we have
`0.01624"min"^(-1)=(2.303)/(100" min")"log"(0.4"mol L"^(-1))/([R]"mol L"^(-1)) or " log"(0.4)/([R]) = (0.01624xx100)/(2.303)=0.7052`
Taking antilogs, we have
`(0.4)/([R])=5.072 or [R]=(0.4)/(5.072)=0.079"mol L"^(-1)`
(iii) Initial rate of reaction
`=((0.400 "mol L"^(-1)-0.289"mol L"^(-1)))/(20" min")=(0.111" mol L"^(-1))/(0" min")=0.00555" mol L"^(-1)"min"^(-1)`