Areaction is followed and found to have a rate constant of 3.36 × 104 m-1s-1 at 344 k and a rate constant of 7.69 m-1s-1 at 219 k. determine the activation energy for this reaction.23.8 kj/mol11.5 kj/mol12.5 kj/mol42.0 kj/mol58.2 kj/mol

Ea = 42,0 kJ/mol

Explanation:

It is possible to solve this question using Arrhenius formula:

Where:

k1: 3,36x10⁴ M⁻¹ s⁻¹

T1: 344K

Ea = ???

R = 8,314472x10⁻³ kJ/molK

k2 : 7,69 M ⁻¹ s⁻¹

T2: 219K

Solving:

Ea = 42,0 kJ/mol

I hope it helps!

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

58750.08068 is the answer i think