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For the non-equilibrium process, A + B r...

For the non-equilibrium process, `A + B rarr` Product, the rate is first-order w.r.t. `A` and second order w.r.t. `B`. If `1.0 mol` each of `A` and `B` were inrofuced into `1.0 L` vessel and the initial rate was `1.0 xx 10^(-2) mol L^(-1) s^(-1)`, calculate the rate when half the reactants have been turned into Products.

A

`1.2xx10^(-3)`

B

`1.2xx10^(-2)`

C

`2.5xx10^(-4)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
a
Rate `=K[A][B]^(2)`
`:. 10^(-2)=K[1][1]^(2)`
or `K=10^(-2) litre^(2) mol^(-2) sec^(-1)`
New rate `=10^(-2)xx0.5xx(0.5)^(2)`
`=1.2xx10^(-3) mol litre^(-1) sec^(-1)`
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