PartC The reactant concentration in a first-order reaction was 7.80x102 Mafter 1

Question

PartC The reactant concentration in a first-order reaction was 7.80x102 Mafter 15.0 s and 6.30x103 M after 65.0s. What is the rate constant for this reaction? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash. View Available Hint(s) kist-0.042 S SubmitPrey X Incorrect; Try Again; 4 attempts remaining Previous Answers Part D The reactant concentration in a second-order reaction was 0.240 M after 220 s and 3.40x102 M after 850 s. What is the rate constant for this reaction? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a daralh View Available Hint(s) k2ndValue Units Submit

Explanation / Answer

Part c

For a first order reaction

The integrated rate law

[A] = [A0] exp(-kt)

[A] = concentration of A after time t

[A0] = initial concentration of A

From the first data

t = 15 s

[A] = 7.8*10^-2 M

7.8*10^-2 = [A0] exp(-k*15) ........ Eq1

From the second data

t = 65 s

[A] = 6.3*10^-3 M

6.3*10^-3 = [A0] exp(-k*65) ....... Eq2

Divide eq1/eq2

12.38 = exp(-k*15) / exp(-k*65)

12.38 = exp[(-k*15) - (-k*65)]

2.516 = - 15k + 65k

2.516 = 50k

k = 0.0503 s-1

Part d

For a second order reaction

The integrated rate law

1/[A0] = 1/[A] - kt

For first data

t = 220 s

[A] = 0.240 M

1/[A0] = 1/(0.240) - k*220 .... Eq1

For second data

t = 850 s

[A] = 3.4*10^-2 M

1/[A0] = 1/(3.4*10^-2) - k*850 ..... Eq2

Equal the equation 1 and equation 2

1/(0.240) - k*220 = 1/(3.4*10^-2) - k*850

4.166 - 220k = 29.411 - 850k

25.245 = 630k

k = 0.04007 M-1 s-1

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