Question

Define standard enthalpy of formation . write a chemical equation for the formation of N2O5. answer in 100 to 150 words.

Answer 1

The standard enthalpy of formation is the change in enthalpy when one mole of a substance in the standard state that is 1 atm of pressure and 298.15 K, is formed from its pure elements under the same conditions.

Standard formation refers to the formation of a product from its standard elements. Since N2O5 consists of nitrogen and oxygen we can use them as the two reactants. Enthalpy is a state function so we can add up all the sub-parts to get the overall enthalpy. Hess' Law can be used here.

2( 2NO + O2 -> 2NO ) DeltaH1 
4NO2 + O2 -> 2N2O5 DeltaH2 
2( N2 + O2 -> 2NO ) DeltaH3 

Adding these formulas results in the cancellation of NO and NO2

2N2 + 5O2 -> 2N2O5 

DeltaHformation = 2DeltaH1 + DeltaH2 + 2DeltaH3 
= 2(-114.1kJ) + (-110.2kJ) + 2(180.5kJ) 
DeltaHformation= 22.6 kJ 

Answer 2

The standard enthalpy of formation is the change in enthalpy when one mole of a substance in the standard state that is 1 atm of pressure and 298.15 K, is formed from its pure elements under the same conditions.

Standard formation refers to the formation of a product from its standard elements. Since N2O5 consists of nitrogen and oxygen we can use them as the two reactants. Enthalpy is a state function so we can add up all the sub-parts to get the overall enthalpy. Hess' Law can be used here.

2( 2NO + O2 -> 2NO ) DeltaH1
4NO2 + O2 -> 2N2O5 DeltaH2
2( N2 + O2 -> 2NO ) DeltaH3

Adding these formulas results in the cancellation of NO and NO2

2N2 + 5O2 -> 2N2O5

DeltaHformation = 2DeltaH1 + DeltaH2 + 2DeltaH3
= 2(-114.1kJ) + (-110.2kJ) + 2(180.5kJ)
DeltaHformation= 22.6 kJ