Part #1: A reaction is second order. If the final concentration of [A] is 0.0505

Question

Part #1:

A reaction is second order. If the final concentration of [A] is 0.0505 M in the reaction container, how many seconds have passed from the time when the initial concentration was 0.1590 M? Note, k for the reaction at this temperature  is 6.712 M-1s-1.

Part #2:

In a second experiment, it was determined that a plot of the natural log of the concentration vs time was linear. Determine the initial concentration of the substance 'A' if 2.0 seconds has past and the final concentration is 5.689 M. Note: k = 0.109 s-1.

Explanation / Answer

1)

for 2nd order

1/A = 1/A0 + kt

solve for t

1/0.0505 = 1/0.159 + (6.712) * t

t = (1/0.0505 - 1/0.159 )/(6.712) = 2.0132seconds

2)

if this is linear, then this must be 1st order

A = A0*exp(-kt)

5.689 = A0*exp(-0.109*2

A0 = 5.689 / (exp(-0.109*2) = 7.0747668 M

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