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

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

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?

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

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

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).