For the reaction, `A+Brarr"product"` If concentration of A is doubled, rate increases 4 times. If concentrations of A and B both are doubled, rate inc

Question

For the reaction, `A+Brarr"product"` If concentration of A is doubled, rate increases 4 times. If concentrations of A and B both are doubled, rate inc

For the reaction, `A+Brarr"product"`
If concentration of A is doubled, rate increases 4 times. If concentrations of A and B both are doubled, rate increases 8 times . The differential rate equation of the reaction will be
A. `(dc)/(dt)-kC_(A)xxC_(B)`
B. `(dC)/(dt)="k "C_(A)^(2)xxC_(B)^(3)`
C. `(dC)/(dt)=kC_(A)^(2)xxC_(B)`
D. `(dC)/(dt)=kC_(A)^(2)xxC_(B)^(2)`

Monjur Alahi

9 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - C
Let the order with repect to A and B are x and y, respectively . Hence , Rate , `r=[A]^xx[B]^Y`
On doubling the concentration of A, rate increases 4 times , `4r=[2A]^xx[B]^Y`
From Eqs. (i) and (ii),
`(1)/(4)=((1)/(2))^(x)`
`:." "x=2`
`:.` Order with respect to A is 2.
If concentration of A and B both are doubled,
`8r=[2A]^x[2B]^y`
From Eqs. (i) and (ii), we get
`(1)/(8)=(1)/(2)^(x).(1)/(2)^(y)" "[becausex=2]`
`(1)/(8)=(1)/(4xx2^(y))rArr2^(y)=2`
`:." "y=1`
Hence, the differential rate equation is
`rprop[A]^(2)[B]^(1)" "or(dc)/(dt)=kC_(A)^(2)xxC_(B)`
`[Where, C_(A)andC_(B)=concentrations " of " A andB]`