Part A The reactant concentration in a zero-order reaction was 5.00×102 M after
ID: 978971 • Letter: P
Question
Part A The reactant concentration in a zero-order reaction was 5.00×102 M after 190 s and 3.50×102 M after 370 s . What is the rate constant for this reaction? Express your answer with the appropriate units.
Part B
What was the initial reactant concentration for the reaction described in Part A?
Express your answer with the appropriate units
Part C
The reactant concentration in a first-order reaction was 8.80×102M after 20.0 s and 7.10×103M after 60.0 s. What is the rate constant for this reaction?
Express your answer with the appropriate units.
Part D
The reactant concentration in a second-order reaction was 0.650 M after 190 s and 1.60×102M after 855 s . What is the rate constant for this reaction?
Express your answer with the appropriate units. Include an asterisk to indicate a compound unit with mulitplication, for example write a Newton-meter asN*m.
Explanation / Answer
Part A
zero order: [A]=-kt+[Ao]
[A]= conc. of reactant at time t.
Ao is initial reactant concentration, k is the rate constant.
5 * 10-2 = -k(190)+[Ao]
3.5 * 10-2 = -k(370)+[Ao]
subtracting 5*10-2 - 3.5*10-2 =-190 k -(-370k)
180k=0.015
k=8.33 x 10-5 mole/L*s
Part B
that's the intercept of that line...
[At] = -k x t + [Ao]
[Ao] = [At] + (8.33x10-5 M/s) x t
pick either point...
[Ao] = (5*10-2M) + (8.33x10-5 M/s) x (190s) = 0.065M
Part C
rate = -d[A] / dt = k x [A]¹
rearranging...
1 / [A]¹ d[A] = -k dt
integrating..
ln[At] - ln[Ao] = -kt
rearranging...
ln[At] = -kt + ln[Ao]
also of the form..
y = mx + b
so a plot of t vs ln[At] gives slope = -k and intercept = ln[Ao]
ie...
k = - (ln[A2] - ln[A1]) / (t2 - t1) = - ln([A2] / [A1]) / (t2 - t1)
k = - ln(7.10x10-3 / 8.80x10-2) / (60s - 20.0s)
k = 0.0629 / s...
Part D
rate = -d[A] / dt = k x [A]²
rearranging...
1 / [A]² d[A] = -k dt
integrating..
- 1/[At] - -1/[Ao] = -kt
rearranging...
1/[At] -1/[Ao] = kt
rearranging...
1/[At] = kt + 1/[Ao]
also of the form..
y = mx + b
so a plot of t vs 1/[At] gives slope = k and intercept = 1/[Ao]
ie...
k = + (1 / [A2] - 1 / [A1]) / (t2 - t1)
k = + (1/1.60x10-2M - 1/0.650M) / (855s - 190s)
k = 0.0916 / (Mxsec).