Calc Enthalpy of Formation N2O5(g) from Data

[chemister]

How would you calculate the standard enthalpy of formation of N2O5(g) in kJ/mol from the following data (all species are gases):
2 NO + O2-----2 NO2 ∆H° = –114.1 kJ
4 NO2 + O2-----2 N2O5 ∆H° = –110.2 kJ
N2 + O2------2 NO ∆H° = +180.5 kJ

Any help is greatly appreciated!

 

[DH]

How would you calculate the standard enthalpy of formation of N2O5(g) in kJ/mol from the following data (all species are gases):
2 NO + O2-----2 NO2 ∆H° = –114.1 kJ
4 NO2 + O2-----2 N2O5 ∆H° = –110.2 kJ
N2 + O2------2 NO ∆H° = +180.5 kJ

Any help is greatly appreciated!

ok, here is your answer. this one is the easy one!
time first and third one with 2, then add three of them up, we have:
4 NO + 2O2 + 4NO2 + O2 + 2N2 + 2O2 -> 4NO2 +2N2O5 + 4NO

5O2 + 2N2 -> 2N2O5 ∆H° = 2*(-114.1) -110.2 + 2*180.5 = 22.6
-> standard enthalpy of formation of N2O5 = 22.6/2 = 11.3 kJ

 

[kashmiri]

heat of formation for lactic acid

 

[kashmiri]

what is the standard value of heat of formation for lactic acid

 

[LudusRex]

To calculate the standard enthalpy of formation of N2O5(g), we can use the following equation:

ΔH°f(N2O5) = ΣnΔH°f(products) - ΣnΔH°f(reactants)

First, we need to determine the number of moles (n) for each species in the reaction. From the given data, we can see that 2 moles of NO and 1 mole of O2 react to form 2 moles of NO2, which has a ΔH° of -114.1 kJ. This means that the number of moles for NO and O2 is equal to 2 and 1, respectively.

Next, 4 moles of NO2 and 1 mole of O2 react to form 2 moles of N2O5, which has a ΔH° of -110.2 kJ. This means that the number of moles for NO2 and O2 is equal to 4 and 1, respectively.

Finally, 1 mole of N2 and 1 mole of O2 react to form 2 moles of NO, which has a ΔH° of +180.5 kJ. This means that the number of moles for N2 and O2 is equal to 1 and 1, respectively.

Now, we can plug in these values into the equation to calculate the standard enthalpy of formation of N2O5:

ΔH°f(N2O5) = (2 mol)(-110.2 kJ/mol) + (4 mol)(-114.1 kJ/mol) - (1 mol)(+180.5 kJ/mol) = -218.5 kJ/mol

Therefore, the standard enthalpy of formation of N2O5(g) is -218.5 kJ/mol.