- #1
jtruth914
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A second- order reaction of the type A + B -->P was carried out in a solution that was initially 0.075 mol dm^-3 in A and 0.050 mol dm^-3 in B. After 1.0 h the concentration of A had fallen to 0.020 mol dm^-3. a) Calculate the Rate constant. b) Solve for the half- life of each of the reactants.
Hint: Answers are a) 16.22 dm^3/mol*h
b) 5.1 × 10^3 s, 2.1 × 10^3 s
My Attempt:
[A]intial= 0.075 mol dm^-3
[A]= 0.20
x=[A]initial-[A]=0.075-0.20= 0.055
intial=0.050 mol dm^-3
=0.050-x= 0.050-0.055= -0.005
this is where I'm confused because I get a negative for . When I plug that into the integrated rate law, I get a nonreal number. The integrated rate law is
kt(initial-[A]initial)=ln([A]intial/[A]intial)
Any idea?
Hint: Answers are a) 16.22 dm^3/mol*h
b) 5.1 × 10^3 s, 2.1 × 10^3 s
My Attempt:
[A]intial= 0.075 mol dm^-3
[A]= 0.20
x=[A]initial-[A]=0.075-0.20= 0.055
intial=0.050 mol dm^-3
=0.050-x= 0.050-0.055= -0.005
this is where I'm confused because I get a negative for . When I plug that into the integrated rate law, I get a nonreal number. The integrated rate law is
kt(initial-[A]initial)=ln([A]intial/[A]intial)
Any idea?