Could someone please help me on part a? Thanks!
For the first-order reaction A-->3B + C when [A]0=0.015 mol.L^-1, the concentration of B increases to 0.018 mol.L^-1 in 3.0 min
a) What is the rate constant for the reaction expressed at the rate of loss of A?
7B.9a
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Lexie Baughman 2C
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Re: 7B.9a
First calculate the concentration of A at 3 minutes.
[A(t)] = [A(0)] - (1 mol A/3 mol B) x [B(t)] = .015mol/L - (1/3) x .018 mol/L = .009 mol/L
The rate constant is then determined from the first-order integrated rate law.
k = (ln([A(0)]/[A(t)]) / t = ln (.015/.009) / 3 min = .17 min^-1
[A(t)] = [A(0)] - (1 mol A/3 mol B) x [B(t)] = .015mol/L - (1/3) x .018 mol/L = .009 mol/L
The rate constant is then determined from the first-order integrated rate law.
k = (ln([A(0)]/[A(t)]) / t = ln (.015/.009) / 3 min = .17 min^-1
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