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InterviewSolution
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The following rate data were obtained at `303 K` for the following reaction: `2A + B rarr C + D` What is the rate law? What is the order with respect to each reactant and the overall order? Also calculate the rate constant and write its units. |
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Answer» Let the order of the reaction with respect to A be x and with respect to B be y . Therefore , rate of the reaction is given by , Rate = `k[A]^(x) [B]^(y)` According to the question , `6.0 xx 10^(-3) = k [0.1]^(x) [0.1]^(y) " " (i)` `7.2 xx 10^(-2) = k [0.3]^(x) [0.2]^(y) " " (ii)` `2.88 xx 10^(-1) = k [0.3]^(x) [0.4]^(y) " " (iii) ` `2.40 xx 10^(-2) = k [0.4]^(x) [0.1]^(y) " " (iv)` Dividing equation (iv) by (i) , we obtain `(2.40 xx 10^(-2))/(6.0 xx 10^(-3)) = (k [0.4]^(x)[0.1]^(y))/(k [0.1]^(x) [0.1]^(y))` `implies 4 = ([0.4]^(x))/([0.1]^(x))` `implies 4 = ((0.4)/(0.1))^(x)` `implies (4)^(1) = 4^(x)` `implies x = 1 ` Dividing equation (iii) by (ii) , we obtain `(2.88 xx 10^(-1))/(7.2 xx 10^(-2)) = (k [0.3]^(x) [ 0.4]^(y))/(k[0.3]^(x) [ 0.2]^(y))` `implies 4 = ((0.4)/(0.2))^(y)` `implies 4 = 2^(y)` `implies 2^(2) = 2^(y)` `implies y = 2 ` Therefore , the rate law is Rate = k `[A][B]^(2)` `implies k = ("Rate")/([A][B]^(2))` From experiment l , we obtain `k = (6.0 xx 10^(-3) "mol" L^(-1) "min"^(-1))/((0.1 "mol" L^(-1)) (0.1 "mol" L^(-1))^(2))` `6.0 L^(2) "mol"^(-2) "min"^(-1)` From experiment lll , we obtain `k = (2.88 xx 10^(-1) "mol" L^(-1) "min"^(-1))/((0.3 "mol" L^(-1)) (0.4 "mol" L^(-1))^(2))` `6.0 L^(2) mol^(-2) "min"^(-1)` From experiment IV , we obtain `k = (2 . 40 xx 10^(-2) "mol" L^(-1) "min"^(-1))/((0.4 "mol" L^(-1)) (0.1 "mol" L^(-1))^(2))` `= 6.0 L^(2) mol^(-2) "min"^(-1)` Therefore , rate constant , K = `6.0 L^(2) mol^(-2) "min"^(-1)` |
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