a compound has a percent compostion of carbon equal to 48.8383%, hydrogen equal to 8.1636%, and oxygen equal to 43.1981%. what i
1 answer:
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Answer:
<u>D. It will decrease by a factor of 4</u>
Explanation:
According to the question , the equation follows :
Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.
.................(1)
STEP": First, find out the power "a" and "b"
a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)
According to question, <u><em>doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,</em></u>
r' = 2r if [A'] = 2[A]
Here [B] is uneffected means [B']=[B]
hence new rate =
Put the value of [A'] , [B'] and r' in the above equation:
...........(2)
Divide equation (1) by (2) we , get
Here A and A cancel each other
B and B cancel each other
We get,
1^b = 1 ( power of 1 = 1)
This is possible only when a = 1
We know that : a + b = 3
1 + b = 3
b =3 -1 = 2
b = 2
Hence the rate law becomes :
<u>.............(3)</u>
Look in the question now, it is asked to calculate the concentration of [B],if cut in half
Hence
[B']=1/2[B]
Insert the value of [B'] in equation (3)
............(a)
But
..............(b)
Compare equation (a) and (b) , we get
new rate r' =
<u>r' = 1/4 r</u>