7B.9a


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Xuan Kuang 2L
Posts: 31
Joined: Wed Nov 14, 2018 12:23 am

7B.9a

Postby Xuan Kuang 2L » Sun Mar 10, 2019 7:21 pm

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?

Lexie Baughman 2C
Posts: 30
Joined: Sat Oct 06, 2018 12:16 am

Re: 7B.9a

Postby Lexie Baughman 2C » Sun Mar 10, 2019 7:25 pm

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


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