Question 14.103

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

Justin Le 2I
Posts: 142
Joined: Fri Sep 26, 2014 2:02 pm

Question 14.103

The only part of this question I do not understand is part f. Can anyone explain please?

Regina Chi 2K
Posts: 51
Joined: Fri Sep 26, 2014 2:02 pm

Re: Question 14.103

Hm... I don't really remember exactly why this is true and don't know if I am able to explain it clearly, but if you look in 14.3 Rate Laws and Reaction Order, you would see that the initial rate of consumption = k [x]initial...

And they noted that if you plot the concentration against the initial rate, you should get a straight line. This is probably because initial rate of consumption = k [x]initial resembles y = mx + b where y = initial rate of consumption, k = m, [x]initial = x, and b = 0...

Therefore, you would get a straight line plot. I hope this helped!