(Hw prob 14. 23) Finding rate constant for 1st order rxn

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Mia-Kyla Coronel 2N
Posts: 7
Joined: Fri Sep 26, 2014 2:02 pm

(Hw prob 14. 23) Finding rate constant for 1st order rxn

Postby Mia-Kyla Coronel 2N » Sun Feb 22, 2015 3:54 pm

Part C
2A-->B + C, given that [A]o=0.153 mol.L^-1, and that after 115 s the concentration of B rises to 0.034 mol*L^-1.

In the solutions manual:
[A]=(0.153 mol A/L)-[(2molA/1molB)*(0.034 mol B/L)]=0.085 (mol A)*L^-1

this probably has a simple answer, but why do we subtract the two?

thanks in advance!

Niharika Reddy 1D
Posts: 127
Joined: Fri Sep 26, 2014 2:02 pm

Re: (Hw prob 14. 23) Finding rate constant for 1st order rxn

Postby Niharika Reddy 1D » Sun Feb 22, 2015 3:56 pm

We subtract them to see how much of the initial has been used up and what concentration is left after producing that much of the product B.

Neil DSilva 1L
Posts: 70
Joined: Fri Sep 26, 2014 2:02 pm

Re: (Hw prob 14. 23) Finding rate constant for 1st order rxn

Postby Neil DSilva 1L » Sun Feb 22, 2015 4:01 pm

That's just the initial concentration of A subtracted by the change in A.

The change in A is stoichiometrically determined using the change in B (A decreases twice as fast as B increased).


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