2N2O+O2=4NO calculate enthalpy
dHrxn = (n*dHo products) - (n*dHo reactants)
YOu have tables in your text. If you don't have a text use the web to find dH of and calculate dHrxn.
To calculate the enthalpy change for a reaction, you would need the standard enthalpy of formation (∆Hf°) values for each compound involved in the reaction. The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states at a specified temperature and pressure (usually 298 K and 1 atm).
Here are the standard enthalpy of formation values for the compounds in your reaction:
- N2O: +82.05 kJ/mol
- O2: 0 kJ/mol
- NO: +90.4 kJ/mol
Given these values, you can calculate the enthalpy change using the following equation:
∆Hrxn° = ∑(n∆Hf° products) - ∑(n∆Hf° reactants)
Where ∆Hrxn° is the enthalpy change of the reaction, n is the stoichiometric coefficient for each compound.
In your reaction:
Reactants:
- 2N2O
- 1O2
Products:
- 4NO
Plugging in the values:
∆Hrxn° = [4(90.4 kJ/mol)] - [2(82.05 kJ/mol) + 1(0 kJ/mol)]
Simplifying the equation:
∆Hrxn° = 361.6 kJ/mol - 164.1 kJ/mol
∆Hrxn° = 197.5 kJ/mol
Therefore, the enthalpy change (∆Hrxn°) for the given reaction is +197.5 kJ/mol.