## How to calculate [A]t in 7B.3 7th ed

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

JulieAljamal1E
Posts: 71
Joined: Fri Sep 28, 2018 12:24 am

### How to calculate [A]t in 7B.3 7th ed

In part C, we’re asked to calculate the rate constant for the rate loss of A from 2A->B+C given that [A]0=0.153 mol/L and that after 115s give the the concentration of B rises to 0.034 mol/L. To do this, I know we have to uses ln([A]t/[A]0)=-kt. But first we need to come of with [A]t. The solutions shows that [A]t=(0.153molA/L)-[(2molA/1molB)(0.034molB/L)=0.085. I dont understand why we have to subtract (2/1)(.034) from 0.153. That makes it seem like we’re calculating the change in concentration of A. Why is [A]t not just (2molA/1molB)(0.034molB/L)??

Shubham Rai 2C
Posts: 64
Joined: Fri Sep 28, 2018 12:27 am

### Re: How to calculate [A]t in 7B.3 7th ed

What you are doing is finding the leftover [A] because if you just do (2molA/1molB)(0.034molB/L), it gives you the change in [A] based on how much the [B] changed. That's why you have to subtract the change from the initial to get the [A]t.