The answer would be A) cells!!
Answer:
By a factor of 12
Explanation:
For the reaction;
A + 2B → products
The rate law is;
rate = k[A]²[B]
As you can see, the rate is proportional to the square of the concentration of A and the of the concentration of B
.
Let's say initially, [A] = x, [B] = y
The rate law in this case is equal to;
rate1 = k. x².y
Now you double the concentration of A and triple the concentration of B.
[A] = 2x, [B] = 3y
The new rate law is given as;
rate2 = k . (2x)². (3y)
rate2 = k . 4x² . 3y
rate2 = 12 k . x² . y
Comparing rate 2 and rate 1, the ratio is given as; rate 2/ rate 1 = 12
Therefore the rate has increased by a factor of 12.
Answer:
C) In[reactant] vs. time
Explanation:
For a first order reaction the integrated rate law equation is:
where A(0) = initial concentration of the reactant
A = concentration after time 't'
k = rate constant
Taking ln on both sides gives:
![ln[A] = ln[A]_{0}-kt](https://tex.z-dn.net/?f=ln%5BA%5D%20%3D%20ln%5BA%5D_%7B0%7D-kt)
Therefore a plot of ln[A] vs t should give a straight line with a slope = -k
Hence, ln[reactant] vs time should be plotted for a first order reaction.
