A first-order reaction initially proceeds at a rate of 0.5 mol/ls. What will be the rate when half the starting material remains? When one-fourth of the starting material remains?
im lost as hell
You let A to be any thing, any thing what?
I think that is good cause the rate is proportional to [A] as it is first order?
Well, isn't that a reaction rate that's first in line? It's moving at a speed of 0.5 mol/ls initially, but what about when half of the starting material remains? It's like a traffic jam where some cars have already left the party. The rate at that point will be half of the initial rate, so it'll be 0.25 mol/ls.
And what about when only one-fourth of the starting material remains? Well, it's like a shrinking crowd at a clown convention. The rate at that point will be one-fourth of the initial rate, so it'll be 0.125 mol/ls. It's all about proportion, my friend.
To answer this question, we need to use the integrated rate law for first-order reactions.
The integrated rate law for a first-order reaction is given by the equation: ln([A]/[A]0) = -kt
Where [A] is the concentration of the reactant at a given time, [A]0 is the initial concentration of the reactant, k is the rate constant, and t is the time.
Let's assume the initial concentration of the reactant is [A]0 and the concentration after a certain time is [A]. We can rearrange the equation to solve for t:
t = - (1/k) * ln([A]/[A]0)
Now, let's calculate the time required for half the starting material to remain:
t1/2 = - (1/k) * ln(1/2) = (1/k) * ln(2)
Similarly, let's calculate the time required for one-fourth of the starting material to remain:
t1/4 = - (1/k) * ln(1/4) = (1/k) * ln(4)
Since the rate of a first-order reaction is proportional to the concentration of the reactant, the rate constant (k) will be the same at any point during the reaction.
Now, let's find the rate when half the starting material remains:
Using the integrated rate law, we know that the rate of the reaction at any time t is given by:
Rate = k * [A]
When half the starting material remains, the concentration [A] is equal to [A]0/2. Therefore, the rate can be calculated as follows:
Rate when half remains = k * ([A]0/2)
Similarly, let's find the rate when one-fourth of the starting material remains:
When one-fourth of the starting material remains, the concentration [A] is equal to [A]0/4. Therefore, the rate can be calculated as follows:
Rate when one-fourth remains = k * ([A]0/4)
By plugging in the known values into these equations, we can calculate the rates when half and one-fourth of the starting material remains.
Isn't the rate for a first order reaction rate = k[A]?.
rate = 0.5
Make A anything and solve for k.
Then use rate = k[A].
you know k and [A] now is 1/2. Then 1/4. solve for rate for each. Check my thinking.