7.4 Hydrogen peroxide decomposes to produce water and oxygen. The equation is:
2H, O2(g) → 2H, O(g) + O2(g)
Using the following values, calculate the heat
change for the reaction and say whether it is
exothermic or endothermic.
Bond energies:
H—H = 436 kJ/mol
00 = 144 kJ/mol
0=0 = 498 kJ/mol
0-H = 464 kJ/mol

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.

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

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