The initial concentration of `N_(2)O_(5)` in the following first order reaction: `N_(2)O_(5)(g) rarr 2NO_(2)(g)+(1)/(2)O_(2)(g)` was `1.24 xx 10^(-2) mol L^(-1)` at `318 K`. The concentration of `N_(2)O_(5)` after `60 min` was `0.20 xx 10^(-2) mol L^(-1)`. Calculate the rate constant of the reaction at `318 K`.
Text Solution
Verified by Experts
For a first order reaction `"log"0[R]_(1)/([R]_(2))=(k(t_(2)-t_(1)))/(2.303)` `k=(2.303)/((t_(2)-t_(1)))"log"[R]_(1)/[R]_(2)` `=(2.303)/((60"min"-0min))"log"(1.24xx10^(2)"molL"^(-1))/(0.20xx10^(2)"molL"^(-1)) ` `=2.303/60"log6.2"mi n^(-1)` `k = 0.0304 "min"^(-1)`
Topper's Solved these Questions
CHEMICAL KINETICS
NCERT|Exercise Exercise|1 Videos
CHEMICAL KINETICS
NCERT|Exercise SOLVED EXAMPLE|1 Videos
BIOMOLECULES
NCERT|Exercise Exercise|33 Videos
CHEMISTRY IN EVERYDAY LIFE
NCERT|Exercise Exercise|32 Videos
Similar Questions
Explore conceptually related problems
For the first order reaction 2N_(2)O_(5)(g) rarr 4NO_(2)(g) + O_(2)(g)
N_(2)O_(5(g)) rarr 2NO_(2(g)) + (1)/(2)O_(2(g)) . In the above first order reaction the initial concentration of N_(2)O_(5) is 2.40 xx 10^(-2) mol L^(-1) at 318 K. The concentration of N_(2)O_(5) after 1 hour was 1.60 xx 10^(-2) mol L^(-1) . The rate constant of the reaction at 318K is ______ xx 10^(-3) "min"^(-1) . (Nearest integer) [Given: log3= 0.477, log 5= 0.699]
The rate constant of the reaction 2H_(2)O_(2)(aq) rarr 2H_(2)O(l) + O_(2)(g) is 3 xx 10^(-3) min^(-1) At what concentration of H_(2)O_(2) , the rate of the reaction will be 2 xx 10^(-4) Ms^(-1) ?
The rate constant for the reaction 2N_(2)O_(5) rarr 4NO_(2)+O_(2) is 3.0 xx 10^(-5) s^(-1) . If the rate is 2.40 xx 10^(-5) mol L^(-1) s^(-1) , then the concentration of N_(2)O_(5) (in mol L^(-1) ) is
The rate constant for the reaction: 2N_(2)O_(5) rarr 4NO_(2)+O_(2) is 3.0xx10^(-5) sec^(-1) . If the rate is 2.40xx10^(-5) M sec^(-1) , then the concentration of N_(2)O_(5) (in M) is:
