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^{−3}M−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^{−4}M/s. If after 45.0 seconds the concentration has dropped to 8.50×10^{−2}M, what was the initial concentration?
Express your answer with the appropriate units.
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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