InterviewSolution
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InterviewSolution
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For the reaction, `2A + B to A_(2)B`, the reaction rate = `k[A][B]^(2)` with k =`2.0 xx 10^(-6) mol^(-2) L^(2)s^(-1)`. Calculate the initial rate of the reaction when [A] = 0.1 mol `L^(-1)`, [B]= 0.2 mol`L^(-1)`. Also calculate the reaction rate when [a] is reduced to 0.06 mol `L^(-1)`. |
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Answer» First case: Rate = `k[A][B]^(2)` =`2.0 xx 10^(-6)mol^(-2)L^(2)s^(-1) xx (0.1 xx mol L^(-1)) xx (0.2 mol L^(-1))^(2)` `8 xx 10^(-9) mol L^(-1)s^(-1) = 8 xx 10^(-9)Ms^(-1)`. Second case: The concentration of A after taking part in the reaction = 0.06 mol `L^(-1)` Amount of B reacted `= (0.1 - 0.06) = 0.04 mol L^(-1)` Amount of B reacted `= 1/2 xx 0.04 mol L^(-1) = 0.02 mol L^(-1)` The concentration of B after taking part in the reaction = `(0.2-0.02)=0.18 mol L^(-1)` Rate = `k[A][B]^(2)` `=2.0 xx 10^(-6) mol^(-2)L^(2)s^(-1) xx (0.06 mol L^(-1)) xx (0.18 mol L^(-1))^(2)` `=3.89 xx 10^(-9)mol L^(-1)s^(-1) = 3.89 xx 10^(-9)Ms^(-1)` |
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