The oxidation of copper(I) oxide, Cu2O(s) , to copper(II) oxide, CuO(s) , is an exothermic process. 2Cu2O(s)+O2(g)⟶4CuO(s) The change in enthalpy upon reaction of 42.60 g Cu2O(s) is −43.47 kJ . Calculate the work, , and energy change, Δrxn , when 42.60 g Cu2O(s) is oxidized at a constant pressure of 1.00 bar and a constant temperature of 25∘ C . Note that ΔErxn is sometimes used as the symbol for energy change instead of Δrxn .

The work is  

The energy  change of the reaction is    

Explanation:

From the question we are told that

    The chemical equation for this reaction is

     The mass of   is  

      The enthalpy is   this also the change in energy in terms of heat

      The pressure at which it is oxidized is

The  no of moles of used in this reaction is mathematically represented as

The molar mass of is  a constant with a value  

   Now substituting values

From the reaction we see that

Two mole of  reacts with One mole of to give Four moles of  

This means that

One mole of    reacts with 0.5 mole of to give two moles of  

it also implies that

of    reacts with moles of to give moles of  

so

of    reacts with moles of to give moles of  

Now the number of moles of gaseous reactant is

The number of moles of gaseous product  is

So the change in number of moles for gaseous  compounds is mathematically evaluated as

Now the workdone for  the compound is mathematically represented as

Where R is the gas constant with a value of

            T is the temperature with a value  

     Substituting values

Generally the internal energy change of the reaction can be represented as

Substituting value

This is the internal energy change on the reaction for 42.60 g of

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.