The correct final answers for the 2 parts given above. just show the working wit

Question

The correct final answers for the 2 parts given above. just show the working with the correct final answers. Thnak you

Question8 a) (4 points) A(aq) > B(aq) is a first order reaction with respect to A(aq). The concentration of A(aq) after 100.0s of reaction is 0.588M. The concentration of A(aq) after another 100.0s (so 200.0s after the beginning of the reaction) is 0.344 M. What is the half-life of this reaction? b) (4 points) A(aq) -> B(aq) + C(aq) is a second order reaction with respect to A(aq). When we double the concentration of A(aq) and we heat the system from 20.0? to 40.0?, we note that the rate of reaction increases by a factor of 20.0 (the reaction goes 20X faster). What is the activation energy, in kJ/mol, for this reaction?

Explanation / Answer

Part a

Initial concentration [A0] = 0.588 M

Final concentration [A] = 0.344 M

Time t = 100 s

Rate constant k =?

For first order reaction

ln [A] /[A0] = - kt

ln (0.344/0.588) = - k*100

-0.536 = - 100k

k = 5.36 x 10^-3 s-1

Half life of first order reaction

t1/2 = 0.6932/k = 0.6932/(5.36 x 10^-3)

= 129.3 s

Part b

Let initial rate = r1

Temperature T1 = 20 + 273 = 293 K

Final rate r2 = 20r1

Temperature T2 = 40 + 273 = 313 K

r1 = k1 [A1]2

r2 = k2 [A2]2

r1/r2 = (k1/k2) [A1/A2]2

r1/20r1 = (k1/k2) [A1/2A1]2

k2/k1 = 5

From the Arrhenius equation

ln (k2/k1) = (Ea/R) (1/T1 - 1/T2)

ln (5) = (Ea/8.314) (1/293 - 1/313)

1.609 = Ea x 2.623 x 10^-5

Ea = 61342 J x 1kJ/1000 J

Ea = 61.34 kJ

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