Question

Calculate the standard enthalpy of formation of NOCl(g) at 25 ºC, knowing that the standard enthalpy of formation of NO(g) at that temperature is 90.25 kJ / mol, and that the standard molar enthalpy of the reaction 2 NOCl(g) → 2 NO(g) + Cl2(g) is 75.5 kJ / mol at the same temperature.

Answer 1

To find the standard enthalpy of formation of NOCl(g), we use the provided enthalpy of formation for NO(g) and the enthalpy of the given reaction. The calculated value is 128 kJ/mol for NOCl(g) at standard conditions.

Step-by-step explanation:

To calculate the standard enthalpy of formation of NOCl(g), we will use the information provided for the enthalpy of formation of NO(g) and the standard molar enthalpy of the reaction given.

The reaction for the formation of NOCl(g) from its elements in their standard states is:

1/2 N2(g) + 1/2 O2(g) + 1/2 Cl2(g) → NOCl(g)

We know the enthalpy of the reaction:

2 NOCl(g) → 2 NO(g) + Cl2(g), ΔH° = 75.5 kJ/mol.

Divide this by 2 to find the enthalpy change per mole of NOCl:

ΔH° per NOCl = 75.5 kJ/mol / 2 = 37.75 kJ/mol.

This represents the enthalpy change from NOCl to NO. To find the formation enthalpy for NOCl, we need to adjust this value with the known enthalpy of formation for NO:

ΔH°fNOCl = ΔH°fNO + ΔH° per NOCl

ΔH°fNOCl = 90.25 kJ/mol + 37.75 kJ/mol

ΔH°fNOCl = 128 kJ/mol

Answer 2

The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol

Step-by-step explanation:

The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:

Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants

In this case, you have: 2 NOCl(g) → 2 NO(g) + Cl₂(g)

So, ΔH=
`2*H_(NO) +H_{Cl_(2) }-2*H_(NOCl)`

Knowing:

• ΔH= 75.5 kJ/mol
• 
  `H_(NO)`= 90.25 kJ/mol
• 
  `H_{Cl_(2) }`= 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound the chlorine Cl₂)
• 
  `H_(NOCl)`=?

Replacing:

75.5 kJ/mol=2* 90.25 kJ/mol + 0 -
`H_(NOCl)`

Solving

-
`H_(NOCl)`=75.5 kJ/mol - 2*90.25 kJ/mol

-
`H_(NOCl)`=-105 kJ/mol

`H_(NOCl)`=105 kJ/mol

The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol