The four isotopes of lead are shown below, each with its percent by mass abundance and the composition of its nucleus. using these data, calculate the approximate atomic mass of lead.

I believe we are given the following data:

82p, 122n, 1.4% 
82p, 124n, 24.1% 
82p, 125n, 22.1% 
82p, 126n, 52.4% 

Where p stands for proton and n for neutrons. We know that the mass number (atomic mass) is equivalent to the sum of p and n, hence the atomic mass of each isotope is:

isotope 1: 82 + 122 = 204 amu

isotope 2: 82 + 124 = 206 amu

isotope 3: 82 + 125 = 207 amu

isotope 4: 82 + 126 = 208 amu

The atomic mass of Lead is the sum of the weighted average atomic mass of the isotopes, that is:

atomic mass Lead = 204 * 0.014 + 206 * 0.241 + 207 * 0.221 + 208 * 0.524

atomic mass Lead = 207.241 amu = 207.241 g/mol 

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

the potential energy of the reactants is 200j.

from the energy diagram, the energy of the product formed is 350j; this means that, this reaction is an endothermic reaction, because it absorbs energy from its environment.