Question

Part A

The reactant concentration in a zero-order reaction was 5.00×10⁻²M after 130 s and 4.00×10⁻²M after 310 s . What is the rate constant for this reaction?

Express your answer with the appropriate units.

k0th =

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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.

[A]0 =

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Part C

The reactant concentration in a first-order reaction was 9.20×10⁻²M after 40.0 s and 8.20×10⁻³M after 95.0 s . What is the rate constant for this reaction?

Express your answer with the appropriate units.

k1st =

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Part D

The reactant concentration in a second-order reaction was 0.130 M after 295 s and 6.60×10⁻²M 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.

k2nd =

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Answer 1

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)