Question

In: Chemistry

Part A The reactant concentration in a zero-order reaction was 5.00×10−2 M after 190 s and...

Part A The reactant concentration in a zero-order reaction was 5.00×10−2 M after 190 s and 3.50×10−2 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 8.80×10−2M after 20.0 s and 7.10×10−3M 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 1.60×10−2M 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.

Solutions

Expert Solution

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).


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