Given the reaction A + B Solution let a and b be the order with respect to A and

Question

Given the reaction A + B

Explanation / Answer

let a and b be the order with respect to A and B rate = [A]^a[B]^b for the 1st observation: (1.0*10^-4) = (1.0)^a(1.0)^b--------------------->(1) for the 2nd observation: (5.0*10^-5) = (1.0)^a(0.5)^b--------------------->(2) for the 3rd observation: (2.5*10^-5) = (0.5)^a(0.5)^b--------------------->(3) (1)/(2) (1.0*10^-4)/(5.0*10^-5) = (1.0/1.0)^a (1.0/0.5)^b taking log on both side log ((1.0*10^-4)/(5.0*10^-5) ) = log ((1.0/1.0)^a (1.0/0.5)^b) 0.301 = a log(1) + blog(1/0.5) 0.301= 0 + b (0.301) b= 1 similarly (2)/(3) (5.0*10^-5)/(2.5*10^-5) =(1.0/0.5)^a (0.5/0.5)^b taking log on both side log ((5.0*10^-5)/(2.5*10^-5)) = log ((1.0/0.5)^a (0.5/0.5)^b) 0.301 = a log(1/0.5) + b log (1) 0.301 = a*(0.301)+b*0 a = 0.301/0.301 a=1 so the rate expression would be rate = K[A][B]

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