Part A The reactant concentration in a zero-order reaction was 5.00×10 2 M after
ID: 925907 • Letter: P
Question
Part A
The reactant concentration in a zero-order reaction was 5.00×102M after 130 s and 4.00×102M after 310 s . What is the rate constant for this reaction?
Express your answer with the appropriate units.
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Part B
What was the initial reactant concentration for the reaction described in Part A?
Express your answer with the appropriate units.
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Part C
The reactant concentration in a first-order reaction was 9.20×102M after 40.0 s and 8.20×103M after 95.0 s . What is the rate constant for this reaction?
Express your answer with the appropriate units.
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Part D
The reactant concentration in a second-order reaction was 0.130 M after 295 s and 6.60×102M after 900 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 as N*m.
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k0th =Explanation / Answer
Part-A: In case of a zero order reaction, the rate is independent of reactant concentration.
Rate = - k x [R]0 = k0x1 = k
=> Rate = - k (rate constant)
Now rate = (final concentration - initial concentration) / (final time - initial time)
= (4.00 x 10-2 M - 5.00 x 10-2 M) / (310 s - 130 s) = - 5.55 x 10-5 M*s-1
=> Rate = - 5.55 x 10-5 M*s-1
Hence rate cnstant, k = - Rate = - ( - 5.55 x 10-5 M*s-1 ) = 5.55 x 10-5 M*s-1 (answer)
Part-B: Let the initial reactant concentration be [R]0
Since for a zero order reaction rate is constant ( = -k), its value remain same for any interval of time.
rate = (final concentration - initial concentration) / (final time - initial time)
=> - 5.55 x 10-5 M*s-1 = (5.00 x 10-2 M - [R]0 ) / (130 s - 0 s)
=> (5.00 x 10-2 M - [R]0) = 130 s x (- 5.55 x 10-5 M*s-1) = - 7.22 x 10-3 M
=> [R]0 = 5.00 x 10-2 M + 7.22 x 10-3 M = 0.0572 M (answer)
Part-C: Applying the formulae for 1st order reaction,
[R]t = [R]0 x(exp)^-kt
=> 9.20x10-2 M = [R]0 x(exp)^-kx40.0s ----- (1)
8.20x10-3 M = [R]0 x(exp)^-kx95.0s ----- (2)
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Dividing eqn(1) by (2) we get
11.22 = (exp)^55.0k
=> 55.0sk = ln(11.22)
=> k = ln(11.22) / 55.0 s = 0.0440 s-1 (answer)