Which of these species would you expect to have the highest standard entropy (S°)? a. CH4(g)
b. C2H2(g)
c. C2H4(g)
d. C2H6(g)
e. C3H8(g)

e. C₃H₈(g)

Explanation:

Standard entropy is the absolute  entropy of a substance at 1 atm and 25°C.

For  different substances in the same phase, molecular complexity determines which ones have higher entropies.

Consider the gases CH₄, C₂H₂, C₂H₄, C₂H₆, C₃H₈. Propane (C₃H₈) has a more complex structure and hence more ways to execute molecular motions, which also increase its microstates. Therefore,  propane has the greatest standard entropy.

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