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For the reaction : 2A+B rarr A(2)B t...

For the reaction `:`
`2A+B rarr A_(2)B`
the rate `=k[A][B]^(2)` with `k=2.0xx10^(-6)mol^(-2)L^(2)s^(-1)`. Calculate the initial rate of the reaction when `[A]=0.1 mol L^(-),[B]=0.2 mol L^(-1)`. Calculate the rate of reaction after `[A]` is reduced to `0.06 mol L^(-1)`.

Text Solution

Verified by Experts

The initial rate of reaction is
Rate = `k[A][B]^(2)`
= `(2 .0 xx 10^(-6) mol^(-2) L^(2) s^(-1)) 0.1 "mol" L^(-1))^(2)`
=`8.0 xx 10^(-9) mol^(-2) L^(2) s^(-1)`
When [A] is reduced from `0.1` mol `L^(-1)` to `0.06 mol^(-1)` the concentration of A reacted = `(0.1 - 0.06) "mol" L^(-1) = 0.04 "mol" L^(-1)`
Therefore , concentration of B reacted `= (1)/(2) xx 0.04 "mol" L^(-1) = 0.02 "mol" L^(-1)`
Then , concentration of B available , [B] = `(0.2 - 0.02)` mol `L^(-1)`
= `0.18 "mol" L^(-1)`
After [A] is reduced to `0.06` mol `L^(-1)` , the rate of the reaction is given by .
Rate = `k[A][B]^(2)`
= `(2.0 xx 10^(-6) mol^(-2) L^(2) s^(-1)) (0.06 "mol" L^(-1)) (0.18 "mol" L^(-1))^(2)`
= `3.89 "mol" L^(-1) s^(-1)`
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