Part A Calculate the fraction of methyl isonitrile molecules that have an energy

Question

Part A Calculate the fraction of methyl isonitrile molecules that have an energy of 160.0 kJ or greater at 510 K .

Part B Calculate this fraction for a temperature of 524 K .

Part C What is the ratio of the fraction at 524 K to that at 510 K ?

Part A

What is the half-life of a first-order reaction with a rate constant of 7.20×104  s1?

Part B

What is the rate constant of a first-order reaction that takes 2.20 minutes for the reactant concentration to drop to half of its initial value?

Part C

A certain first-order reaction has a rate constant of 3.60×103 s1. How long will it take for the reactant concentration to drop to 18 of its initial value?

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 6.30×103 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?

Express your answer numerically in minutes.

Part B

A certain second-order reaction (Bproducts) has a rate constant of 1.40×103M1s1 at 27 C and an initial half-life of 278 s . What is the concentration of the reactant B after one half-life?

Express the molar concentration numerically.

Explanation / Answer

Part A

For first-order reaction a rate constant k= 7.20×104 s1

For First order reaction, the relation between t½ and rate constant is,

t½ = 0.693/k

t½ = 0.693/(7.20×104)

t½ = 962.5 s

Half life = 962.5 s

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Part B

For a first-order reaction t½ = 2.20 min.

k = 0.693/t½ /

k = 0.693/2.20

k = 0.315 min-1

k = 0.315 / 60 s-1

k = 5.25 x 10-3 s-1

Rate constant k is k = 5.25 x 10-3 s-1

==================

Part C

K = 3.60 x 10-3 s-1

Initial concentration=[A0]

Final concentration at time t [A] = 1/18[A0]

[A] /[A0] = 1/18

Formula,

ln ([A]/[A0]) = -kt

ln (1/18) = -3.60 x 10-3 t

-2.890 = -3.60 x 10-3 t

t= 2.890/-3.60 x 10-3 t

t = 802.88 s

t = 803 s (apprx.)

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2)

Part A

For First order reaction,

K = 6.30×103 s1

At any time t [A] =6.25% of [A0]

So, [A]/[A0] = 6.25 % = 6.25/100

Formula,

ln([A]/[A0]) = -kt

ln(6.25/100) = -6.30×103 x t

-2.7726 = -6.30×103 x t

t = 2.7726/(6.30×103)

t = 440 s

t = 440/60 min

t = 7.33 min.

It will take 7.33 min to rich the concentration to 6.25 % of original.

==================================

Part B

For Second order reaction,

Rate constant K = 1.40 x 10-3 m-1.s-1.

Half life t½ = 278 s

For second order reaction Rate constant(k), Half life (t½) and initial concentration [A0] is given as

t½ = 1/k[A0]

[A0] = 1/ (k t½)

[A0] = 1/(278 x 1.40 x 10-3)

[A0] = 1/0.3892

[A0] = 2.569

i.e. Initial concentration = 2.569

After a half as per definition concentration is half of the initial concentration i.e. ½ [A0]

After one half life,

[A] = ½ [A0] = ½ (2.569)

[A] = 1.2845.

Concentration of B after 1 half life is 1.2845 unit.

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