Consider the reaction: A ---> B
The rate of the reaction is 1.6 x 10-2 M/s when the concentration of A is 0.35 M. Calculate the rate constant if the reaction is (a) first order in A and (b) second order in A.
Answers
Answer:
a)4×10^-2 b)1.3×10^-1
Explanation:
a)R=k[A]
k=R/[A]
b)R=k[A]^2
k=R/[A]^2
Answer:
This equation is of the form y = mx + b, where y = 1/rate, x = [A], m = 1/k, and b = 0. Therefore, plotting 1/rate versus [A] will give a straight line with slope equal to 1
Explanation:
To calculate the rate constant, we need to use the rate law equation for the reaction, which is determined experimentally. The general form of the rate law equation for a chemical reaction is:
rate = k [reactant 1]n [reactant 2]m
where k is the rate constant, n and m are the reaction orders with respect to reactant 1 and reactant 2 respectively.
In this case, we have the reaction:
A ---> B
The rate law equation for this reaction can be written as:
rate = k [A]n
where k is the rate constant and n is the reaction order with respect to A.
(a) First order in A:
If the reaction is first order in A, then the rate law equation becomes:
rate = k [A]
Taking the natural logarithm of both sides gives:
ln(rate) = ln(k) + ln([A])
This is the equation of a straight line with slope equal to the rate constant, k, and y-intercept equal to ln(k).
To determine the rate constant, we need to plot ln(rate) versus time or [A] versus time and find the slope of the linear fit. However, we are given the rate of the reaction, not the concentration of A as a function of time. Therefore, we need to use an alternative approach.
The rate of the reaction, given as 1.6 x 10-2 M/s, is the change in concentration of A per unit time. Therefore, we can write:
rate = -Δ[A]/Δt
where Δ[A] is the change in concentration of A over a certain time interval, and Δt is the time interval.
Substituting the given values, we get:
1.6 x 10-2 M/s = -Δ[A]/Δt
Rearranging the equation gives:
Δ[A]/Δt = -1.6 x 10-2 M/s
The negative sign indicates that the concentration of A decreases with time.
If we assume that the rate is constant over the time interval, we can write:
rate = k [A]
where [A] is the average concentration of A over the time interval. The average concentration can be calculated as:
[A] = (initial concentration + final concentration)/2
Substituting the given values, we get:
[A] = (0.35 M + (0.35 M - 1.6 x 10-2 M/s x Δt))/2
Simplifying the expression, we get:
[A] = 0.35 M - 8 x 10-3 M/s x Δt
Substituting this expression into the rate law equation gives:
1.6 x 10-2 M/s = k (0.35 M - 8 x 10-3 M/s x Δt)
Solving for k gives:
k = 1.6 x 10-2 M/s / (0.35 M - 8 x 10-3 M/s x Δt)
Note that the rate constant depends on the concentration of A at the beginning of the time interval and the duration of the time interval.
(b) Second order in A:
If the reaction is second order in A, then the rate law equation becomes:
rate = k [A]2
Taking the inverse of both sides gives:
1/rate = 1/k [A]2
This equation is of the form y = mx + b, where y = 1/rate, x = [A], m = 1/k, and b = 0. Therefore, plotting 1/rate versus [A] will give a straight line with slope equal to 1
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