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The experiment data for the reaction 2A ...

The experiment data for the reaction `2A + B_(2) rarr 2AB` is
`|{:("Experiment",[A] M,[B_(2)] M,"Initial rate" (mol L^(-1) s^(-1)),),(I,0.50,0.5,1.6 xx 10^(-4),),(II,0.50,1.0,3.2 xx 10^(-4),),(III,1.00,1.0,3.2 xx 10^(-4),):}|`
Write the most probable rate equation for the reacting giving reason for you answer.

Text Solution

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First methof
`2A + B_(2) rarr 2AB`
In experiments `I and II`, the concentration of `A` is constant, but the concentration of `[B_(2)]`is doubled and the rate is also doubled
`:. r prop [B_(2)]^(1)`
ismilarly, in experiments `II and III`, and the concentration of `[B_(2)]` is constant, but the concentration of `[A]` is doubled, but the rate of does not change.
`:. r prop [A]^(0)`
`:.` Rate law is
`r prop [B_(2)][A]^(0)`
or `r = k[B_(2)]`
Second methof
Let the rate of reaction is
`r = k[A]^(alpha)[B_(2)]^(beta)`
`:. r_(1) = 1.6 xx 10^(-4) = k[0.5]^(alpha)[0.5]^(beta)` ...(i)
`r_(2) = 3.2 xx 10^(-4) = k[0.5]^(alpha)[1.0]^(beta)` ...(ii)
`r_(3) = 3.2 xx 10^(-4) = k[1.0]^(alpha)[1.0]^(beta)` ... (iii)
Divide equation (ii) by (i),
`(3.2 xx 10^(-4))/(1.6 xx 10^(-4)) = (k[0.5]^(alpha)[1.0]^(beta))/(k[0.5]^(alpha)[0.5]^(beta))`
`(2)^(1) = (2)^(beta) rArr beta = 1`
ismilarly, divide Eq. (iii) by (ii),
`(3.2 xx 10^(-4))/(3.2 xx 10^(-4)) = (k[1.0]^(alpha)[1.0]^(beta))/(k[0.5]^(alpha)[1.0]^(beta))`
`1 = (2)^(alpha) rArr (2)^(0) = (2)^(alpha) rArr alpha = 0`
`:.` Rate `= k[A]^(0)[B_(2)]^(1) = k[B_(2)]`
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