Question

Part A A certain first-order reaction (A→products) has a rate constant of 7.20×10−3 s−1 at 45...

Part A

A certain first-order reaction (A→products) has a rate constant of 7.20×10−3 s−1 at 45 ∘C. How many minutes does it take for the concentration of the reactant, [A], to drop to 6.25% of the original concentration?

Express your answer with the appropriate units.

Answer:

6.42 min

Part B

A certain second-order reaction (B→products) has a rate constant of 1.35×10−3M−1⋅s−1 at 27 ∘Cand an initial half-life of 236 s . What is the concentration of the reactant B after one half-life?

Express your answer with the appropriate units.

Part C

The rate constant for a certain reaction is k = 2.60×10−3 s−1 . If the initial reactant concentration was 0.100 M, what will the concentration be after 15.0 minutes?

Express your answer with the appropriate units.

Part D

A zero-order reaction has a constant rate of 2.70×10−4M/s. If after 45.0 seconds the concentration has dropped to 8.50×10−2M, what was the initial concentration?

Express your answer with the appropriate units.

Homework Answers

Answer #1

Part A

For the first order reaction

ln(Ao/At) = kt

ln(100/6.25) = 7.20 * 10^(-3) * t

t = 1000 * ln(16)/7.20 = 385.08 seconds

Part B

1/At = 1/Ao + kt

For the first half life

2/Ao = 1/Ao + kt

Ao = 1/kt = 1/(1.35 * 10^(-3) * 236) = 3.183 M

After one half life, concentration of B left will be Ao/2 = 1.569 M

Part C

ln(Ao/At) = kt

ln(Ao/At) = 2.60 * 10^(-3) * 15 * 60

ln(Ao/At) = 2.340

At = 0.100/e^(2.340) = 0.00963 M

Part D

[Ao] = [At] + kt

[Ao] = 8.50 * 10^(-2) + 2.70 * 10^(-4) * 45

[Ao] = 8.50 * 10^(-2) + 1.215 * 10^(-2) = 9.715 * 10^(-2) M

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