15.23

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

donnanguyen1d
Posts: 64
Joined: Fri Sep 29, 2017 7:04 am

15.23

For Part C, why did you have to subtract (2(.034molB)) from (.153 molA) to get [A]t, I thought calculating [A]t would only require multiplying .034molB by 2?

Deborah Cheng 1F
Posts: 50
Joined: Thu Jul 13, 2017 3:00 am

Re: 15.23

I believe it's because the equation of the reaction given is 2A $\rightarrow$ B + C so we need to account for the ratio of 2 mol A used to 1 mol B produced.

Samira 2B
Posts: 38
Joined: Fri Sep 29, 2017 7:05 am

Re: 15.23

It is because that accounts for the change in reactants going to products. Since, products is produce that means the reactants must be reduced. So, you would subtract the 0.068 from the initial reactant concentration.

ConnorThomas2E
Posts: 57
Joined: Fri Sep 29, 2017 7:04 am

Re: 15.23

You start with the initial concentration of A, and you need to find how much b is produced to see how much the concentration of A decreases. Therefore, you would subtract 2*b from the initial concentration of A

Kyra LeRoy 1E
Posts: 52
Joined: Sat Jul 22, 2017 3:00 am

Re: 15.23

For product to form, there must be something taken from the reactant. Therefore, you take the initial concentration of A and subtract the concentration that has been used to create B at time t.