QUESTION 3
Kinetic results for a reaction involving substance A are shown below.
Time (mins)
[A] in mol L-1
0.000
1.00
2.00
0.82
4.00
0.67
7.00
0.49
10.0
0.37
14.0
0.24
20.0
0.14
25.0
0.08
(a) Use Excel to plot and print a graph of these results. (2)
(b) What is the order of this reaction with respect to A? (1)
(c) Use your graph to calculate the half-life for this reaction. (2)
(d) Given that in this reaction, A reacts with G, and that the order with respect to G is second, write the rate equation for this reaction. (2)

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

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.

the answer is a for sure mark brainliest i had got this right on usatestprep