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.
Text Solution
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Rate `=K[A][B]^(2)` `therefore 10^(-2)=K[1][1]^(2)` or `K=10^(-2)" litre"^(2)" mol"^(-2)" sec"^(-1)` Now `"rate"_(a)=10^(-2)xx0.5xx(0.5)^(2)` or New rate `=1.2xx10^(-3)" mol/L-sec"`
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