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The following data were obtained for the...

The following data were obtained for the reaction:
`2NO+O_(2) to 2NO_(2)`

(a) Find the order of reaction with respect to NO and `O_(2)`.
(b) Write the rate law and overall order of reaction.
(c ) Calculate the rate constant (k).

Text Solution

Verified by Experts

(a) Let the order w.r.t. NO and `O_(2)` be x and y.
Rate `=k[NO]^(x)[O_(2)]^(y)`
`7.2xx10^(-2)=k[0.3]^(x)[0.2]^(y) " "…(i)`
`6.0xx10^(-3)=k[0.1]^(x)[0.1]^(y) " "…(ii)`
`2.88xx10^(-1)=k[0.3]^(x)[0.4]^(y) " "...(iii)`
`2.40xx10^(-2) =k[0.4]^(x)[0.1]^(y)" "...(iv)`
Dividing eq. (iv) by eq. (ii), we have
`(2.40xx10^(-2))/(6.0xx10^(-3))=(k[0.4]^(x)[0.1]^(x))/(k[0.1]^(x)[0.1]^(y))`
or `4^(x)=4 or x =1`
Dividing eq. (iii) by eq. (i), we have
`(2.88xx10^(-1))/(7.2xx10^(-2))=(k[0.3]^(x)[0.4]^(y))/(k[0.3]^(x)[0.2]^(y))`
or `2^(y)=4 or y=2`
Order of reaction w.r.t. NO = 1, order of reaction w.r.t. `O_(2)=2`
(b) Rate law:
Rate `=k[NO][O_(2)]^(2)`
Overall order of reaction `=1+2=3`
(c ) Rate constant, `k=("Rate")/([NO][O_(2)]^(2))=(7.2xx10^(-2))/(0.3xx(0.2)^(2))`
or `k=6.0"mol"^(-2)L^(2)"min"^(-1)`
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