The integrated rate laws for zero-, first-, and second-order reaction may be arr

Question

The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they resemble the equation for a straight line,y=mx+b.

Part A) The reactant concentration in a zero-order reaction was 7.00×102M after 105 s and 4.00×102Mafter 390 s . What is the rate constant for this reaction? (show units)

Part B) What was the initial reactant concentration for the reaction described in Part A? (units)

Part C) The reactant concentration in a first-order reaction was 6.80×102M after 35.0 s and 1.20×103Mafter 90.0 s . What is the rate constant for this reaction? (units)

Part D) The reactant concentration in a second-order reaction was 0.230 M after 245 s and 1.50×102M after 820 s . What is the rate constant for this reaction? (units)

Explanation / Answer

part A

rate = dx/dt = x2-x1/t2-t1 = (0.07-0.04)/(390-105) = 1.05*10^-4 M.s-1

for zero order, rate = k

k = rate constant = 1.05*10^-4 M.s-1

part B

for zero order , K = a - (a-x)/t

     (1.05*10^-4)= (a-4*10^-2)/390

a = initial concentration of reactants = 0.081 M

part C

rate constant (k) = (1/t)ln(a/a-x)

at t = 35 sec , a-x = 6.8*10^-2

at t = 90 sec , a-x = 1.2*10^-3

so that,

(1/35)ln(a/(6.8*10^-2)) = (1/90)ln(a/(1.2*10^-3))

a = initial concentration = 0.9 M

k = (1/t)ln(a/a-x)

    = (1/35)ln(0.9/0.068)

     = 0.0738 sec-1

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