2NO2(g)⇌N2O4(g). ΔH∘f for N2O4(g) is 9.16 kJ/mol.
The equation you provided represents a chemical reaction involving the gases nitrogen dioxide (NO2) and dinitrogen tetroxide (N2O4). The symbol (g) stands for "gas," indicating that both NO2 and N2O4 are in the gaseous state.
The equation also includes the notation ⇌, which indicates that the reaction is in equilibrium. This means that the forward and reverse reactions are occurring simultaneously at equal rates, resulting in no net change in the amounts of reactants and products.
The value ΔH∘f for N2O4(g) represents the standard enthalpy of formation for N2O4(g). This is the change in enthalpy that occurs when one mole of N2O4(g) is formed from its constituent elements in their standard states at a given temperature (usually 25°C or 298K) and pressure (usually 1 atm). The standard enthalpy of formation is typically measured in units of kilojoules per mole (kJ/mol) and provides insight into the stability and energy content of a compound.
In this case, the standard enthalpy of formation for N2O4(g) is given as 9.16 kJ/mol. This value indicates that when one mole of N2O4(g) is formed from its constituent elements under standard conditions, 9.16 kJ of energy is either absorbed or released.
To calculate the standard enthalpy of formation, you would need to know the standard enthalpies of formation for the individual elements in their standard states. These values can be found in tables or databases of thermodynamic data. By applying Hess's Law, which states that the total enthalpy change of a reaction is independent of the pathway taken, you can sum the enthalpies of formation of the products and subtract the enthalpies of formation of the reactants to determine the net change in enthalpy.
In this case, since you have already been given the value for ΔH∘f of N2O4(g) as 9.16 kJ/mol, you can directly use this value in any calculations involving the enthalpy change of the reaction or in other thermodynamic calculations.
To calculate the standard enthalpy change (ΔH°) for the reaction 2NO2(g) ⇌ N2O4(g), you need to use the enthalpy of formation (ΔH°f) values for NO2(g) and N2O4(g):
The balanced chemical equation for the reaction is:
2NO2(g) ⇌ N2O4(g)
Here are the given enthalpy of formation values:
ΔH°f(NO2(g)) = ?
ΔH°f(N2O4(g)) = 9.16 kJ/mol
To calculate the ΔH°f(NO2(g)), you need to consider that ΔH°f for elements in their standard states is zero. Therefore, you need to break down the formation of NO2(g) into its individual elements:
ΔH°f(NO2(g)) = [ΔH°f(N2O4(g)) - 2ΔH°f(O2(g))] / 2
Since O2(g) is in its standard state, ΔH°f(O2(g)) = 0.
Thus, you can simplify the equation to:
ΔH°f(NO2(g)) = ΔH°f(N2O4(g)) / 2
Plugging in the given value for ΔH°f(N2O4(g)):
ΔH°f(NO2(g)) = 9.16 kJ/mol / 2
ΔH°f(NO2(g)) = 4.58 kJ/mol
Therefore, the enthalpy of formation for NO2(g) is 4.58 kJ/mol.