In a reaction A to products, ,4.40 min after the reaction is started concentration of A is 0.588M. The rate of reaction at this point is rate= -change in concentration of A/change in time=2.2x10^-2Mmin^-1. Assume that this rate remains constant for a short period of time.
A. what is concentration of A 5.00min after the reaction?
B. At what time after the reaction is started will concentration of A =0.565M?
please I need help in understanding how to solve this question. I already tried it myself a couple of times but I keep getting the wrong answer.
Thanks.
To solve this question, we can use the concept of first-order reaction kinetics. The rate equation for a first-order reaction can be written as:
rate = k[A]
Where rate is the rate of reaction, k is the rate constant, and [A] is the concentration of the reactant A.
Given:
rate = -Δ[A]/Δt = 2.2 x 10^(-2) M/min
[A]t=4.40 min = 0.588 M
A. To find the concentration of A at 5.00 min, we need to assume that the rate remains constant for this short period of time. Since the rate is constant, we can use the equation:
rate = k[A]
Substituting the given values, we have:
2.2 x 10^(-2) M/min = k * 0.588 M
Solving for k:
k = (2.2 x 10^(-2) M/min) / 0.588 M
k = 0.0374 min^(-1)
Now we can use the integrated rate equation for a first-order reaction:
[A]t = [A]0 * e^(-kt)
Where [A]t is the concentration at time t, [A]0 is the initial concentration, k is the rate constant, and e is Euler's number (approximately 2.71828).
Substituting the values at t = 5.00 min:
[A]5.00 = [A]0 * e^(-k * 5.00)
[A]5.00 = 0.588 M * e^(-0.0374 min^(-1) * 5.00 min)
Calculating this expression will give you the concentration of A at 5.00 min.
B. To find the time at which the concentration of A is 0.565 M, we can rearrange the integrated rate equation:
t = (1/k) * ln([A]t/[A]0)
Substituting the values:
t = (1/0.0374 min^(-1)) * ln(0.565 M/0.588 M)
Calculating this expression will give you the time at which the concentration of A is 0.565 M.
Make sure to use a calculator for the logarithmic calculations as the natural logarithm function (ln) is involved.
To solve this question, we need to use the information provided such as the initial concentration of A, the rate of the reaction, and the time elapsed.
Let's go through the steps to find the answers to parts A and B of the question:
A. To find the concentration of A 5.00 min after the reaction, we can use the rate of the reaction information provided.
1. We know that the rate of the reaction is constant, which means that every minute the concentration of A decreases by 2.2x10^-2 M.
2. We can calculate the change in concentration by multiplying the rate by the time:
change in concentration = rate × time
= 2.2x10^-2 M/min × (5.00 min - 4.40 min)
The time difference is 5.00 min - 4.40 min because we want to find the concentration after 5.00 min.
3. Plugging in the values:
change in concentration = 2.2x10^-2 M/min × 0.60 min
= 1.32x10^-2 M
4. Subtract the change in concentration from the initial concentration to find the concentration after 5.00 min:
concentration after 5.00 min = 0.588 M - 1.32x10^-2 M
= 0.57568 M (rounding to four decimal places)
Therefore, the concentration of A 5.00 min after the reaction is approximately 0.5757 M.
B. To find the time at which the concentration of A is 0.565 M, we follow a similar approach:
1. We know that the rate of the reaction is constant, so we can use the rate to determine the change in concentration per minute.
2. First, calculate the change in concentration from the initial concentration to the desired concentration:
change in concentration = 0.588 M - 0.565 M
= 0.023 M
3. Now, we can find the time it takes for this change in concentration to occur:
time = change in concentration / rate
Plugging in the values:
time = 0.023 M / 2.2x10^-2 M/min
≈ 1.045 min (rounding to three decimal places)
Therefore, the concentration of A will be 0.565 M approximately 1.045 min after the reaction is started.
By following these steps, you should be able to get the correct answers to parts A and B of the question.
concA=initial+rate*time
concA=.558M+2.2E-2*5 M
= .558M+2.2E-1=.778M