For a first-order reaction, the half-life is constant. It depends only on the ra

Question

For a first-order reaction, the half-life is constant. It depends only on the rate constant k and not on the reactant concentration. It is expressed as

t1/2=0.693k

For a second-order reaction, the half-life depends on the rate constant and the concentration of the reactant and so is expressed as

t1/2=1k[A]0

Part A

A certain first-order reaction (Aproducts) has a rate constant of 9.00×103 s1 at 45 C. How many minutes does it take for the concentration of the reactant, [A], to drop to 6.25% of the original concentration?

Part B

A certain second-order reaction (Bproducts) has a rate constant of 1.05×103M1s1 at 27 C and an initial half-life of 226 s . What is the concentration of the reactant B after one half-life?

Explanation / Answer

Part A :- For a first order reaction rate constant , k = ( 2.303 /t )x log ( Mo / M)

Where

k = rate constant = 9.00x10-3 s-1

Mo = initial mass

M= Mass left after time t = 6.25% of Mo

    = 6.25 Mo / 100

t = time = ?

PLug the values we get t = ( 2.303 /k )x log ( Mo / M)

                                    = (2.303 / 9.00x10-3 ) x log( Mo /(6.25 Mo / 100))

                                    = 308 s

                                    = 308/60 min

                                    = 5.13 min

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Part B :

For a second order reaction rate constant , t 1/2 = 1/(k[Bo])

Where

[Bo] = initial concentration of A = ?

k = rate constant = 1.05 x 10 -3 M -1 s-1

t1/2 = half life = 226

Plug the values we [Bo] = 4.21 M

For a second order reaction rate constant , k = b / ( ( ta(a-b))

Where

a = initial concentration = [Bo] = 4.21 M

a-b = concentration left after time t = ?

t = time = 1 half life = 226 s

Plug the values we get

k = b / ( ( ta(a-b))

1.05×103 = b / ( 226 x4.21 x (4.21-b))

On solving we get b = 2.10 M

So the concentration of B left after one half life is a-b = 4.21 - 2.10 = 2.11 M

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