A reaction of the form aA-----products is second-order with a rate constant of 0.225 L/(mol/s). If the initial concentration of A is 0.293 mol/L, what is the molar concentration of A after 35.6 secs?
To solve this problem, we can use the second-order rate equation:
rate = k[A]^2
where:
rate = rate of reaction
k = rate constant
[A] = concentration of reactant A
We can rearrange the equation to solve for the concentration of A:
[A]^2 = rate / k
Plugging in the given values:
rate = 0.225 L/(mol/s)
k = 0.225 L/(mol/s)
[A]^2 = 0.225 L/(mol/s) / 0.225 L/(mol/s)
[A]^2 = 1 mol
Taking the square root of both sides:
[A] = √(1 mol)
The molar concentration of A is 1 mol/L.
However, this concentration is for the initial time (t=0). To find the concentration of A after 35.6 seconds, we need to use the integrated rate law for a second-order reaction. The integrated rate law for a reaction of the form aA → products is:
1/[A] = kt + 1/[A]0
where:
[A]0 = initial concentration of A
t = time
k = rate constant
Rearranging the equation to solve for [A]:
[A] = 1 / (kt + 1/[A]0)
Plugging in the values:
[A] = 1 / (0.225 L/(mol/s) * 35.6 s + 1/0.293 mol/L)
Calculating:
[A] = 1 / (7.986 mol + 3.414 mol)
Simplifying:
[A] = 1 / 11.4 mol
The molar concentration of A after 35.6 seconds is approximately 0.0877 mol/L.
To find the molar concentration of A after 35.6 seconds, we can use the second-order rate equation:
Rate = k[A]^2
Where:
- Rate is the rate of the reaction
- k is the rate constant
- [A] is the molar concentration of A
First, let's rearrange the equation to solve for [A]:
[A]^2 = Rate / k
[A] = sqrt(Rate / k)
Now, let's substitute the given values into the equation:
[A] = sqrt(0.225 L/(mol/s) / 0.293 mol/L)^2
[A] = sqrt(0.769 / 0.293)^2
[A] = sqrt(2.628)^2
[A] = 1.62 mol/L
Therefore, the molar concentration of A after 35.6 seconds is 1.62 mol/L.
1/[A] =kt+1/[A]0
=(0.225)(35.6)+(1/0.293)
1/[A]=11.42
[A]=0.088mo.l
(1/A) - (1/Ao) = akt
solve for A.