CHEMISTRY TEST 53 POINTS

How much heat should be transferred when 20.1 g of hydrogen bromide is formed during the reaction of hydrogen gas with liquid bromine?
72.80 kJ + H2 + Br2 → 2 HBr

9.1 KJ

Explanation:

First we must find the number of moles of HBr involved;

number of moles of HBr= mass of HBr/ molar mass of HBr

But molar mass of HBr= 80.91 g/mol

Given mass of HBr from the question= 20.1g

Hence;

Number of moles of HBr= 20.1 g/80.91g

Number of moles of HBr= 0.25 moles of HBr

If 2 moles of HBr has a heat of formation of 72.80KJ

Then 0.25 moles of HBr will have a heat of formation of 0.25× 72.80/2= 9.1 KJ

Therefore; 20.1 g of HBr will have a heat of formation of 9.1 KJ

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