This is a continuation of
Rate and Order of Reactions
\ | Differential rate law | Integrated rate law | Linear plot | Half life |
---|---|---|---|---|
first order | - d[A]/dt = k [A] | [A] = [A]_{o} exp (-k t) | ln [A] vs t slope = - k | hl=ln(2)/k |
second order | - d[A]/dt = k [A]^{2} |
1 1 --- - ---- = k t [A] [A]_{o} |
1 --- vs t [A] slope = k | hl=1/k[A]_{o } |
Problem 1
t / min 0 1 2 4 [A]/[M] 0.100 0.0905 0.0819 0.0670Calculate k. (What is the rate constant?)
Solution
A. We can calculate k from any two data points The integrated rate law for 1st order is A = A_{o} exp (- k t) Using the the first two points, 0.0905 = 0.100 exp (- k * 1) - k = ln (0.0905/0.100) = ln(0.905) = -.0998 min^{-1} Using the point when t = 2 0.100 = 0.0819 exp (-k * 2) - k 2 = ln (0.0819/0.100) = ln (0.819) = -0.200 k = 0.100 min^{-1} Using the point when t = 4 0.0670 = 0.100 exp (- k * 4) 4 k = ln 1.49 = 0.400 k = 0.100 min^{-1} B. Two methods to evaluate half life are: (a) half life * k = ln 2 = 0.693 half life = 0.693 / 0.1 = 6.93 min (note the calculation of units) (b) Calculate the time t when [A] = 0.0500 0.0500 = 0.100 exp (-0.100 t) 0.100 * t = ln (0.100/0.0500) gives the same result. C. When t = 5 min [A] = 0.100 exp (-0.100 *5) = 0.100 * 0.6065 = 0.0607 D. When [A] is reduced by 90%, we have 0.01 = 0.1 exp (-0.100 * t) 0.100 * t = ln (10) t = 2.303/0.100 = 23.03 min check: [A] = 0.1 exp (-0.1 * 23.03) = 0.010
Problem 2
Solution:
In fact, many of the following values can be evaluated without using a calculator.
t/min 0 1 2 5 10 20 30 70. [B]/M 1.0 0.909 0.833 0.667 0.50 0.333 0.25 0.125
Discussion:
From the record shown as above,
Calculate k. (What is the rate constant?)
Calculate the half life for [B] = 1.0.
(What is the half life for an initial concentration of 1.0?)
Calculate [B] when t = 40 min.
Calculate t when [B] = 0.100.
(Estimate the time required for 90% of B to polymerize.)
How does the total pressure change with time?
Getting the answers for the above is fun. Please try.
Discussion -
Consider the relationship
Discussion -
The drop in pressure is 101-95 = 6 kPa. Consider the reaction:
2 C_{4}H_{6}(g) = C_{8}H_{12}(g). 12 kPa 6 kPaSince two moles of butadiene combine to give one mole octadiene, the difference in total pressure is the partial pressure of octadiene, 12 kPa of butadiene converted to 6 kPa of octadiene. What is the partial pressure of butadiene?
Discussion -
The reaction is:
Skill -
Evaluate half-life according to conditions. Apply the equation,