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Question

Dinitrogen tetraoxide (N2O4) is known to be used, along with hydrogen, as rocket fuel by NASA since 1950s. It is prepared as follows:
N2 (g) + 2O2 (g) → N2O4 (g)

Calculated, using the following data, what would be the enthalpy of formation of N2O4?.
(i) 2NO2 (g) → N2 (g) + 2O2 (g) △H = –118.7 kJ
(ii) 2NO2 (g) → N2O4 (g) △H = –101.9 kJ

Answer 1

To find the enthalpy of formation of N2O4, we can use the Hess's Law which states that the enthalpy change of a reaction is the same regardless of the number of steps taken to get from reactants to products.

Given:
(i) 2NO2 (g) → N2 (g) + 2O2 (g) △H = -118.7 kJ
(ii) 2NO2 (g) → N2O4 (g) △H = -101.9 kJ

To find the enthalpy of formation of N2O4, we need to determine the enthalpy change when 1 mole of N2O4 is formed from its elements in their standard states.

First, we can reverse reaction (ii) to get the enthalpy change when 1 mole of N2O4 is decomposed into 2 moles of NO2:
N2O4 (g) → 2NO2 (g) △H = +101.9 kJ

Next, we can reverse reaction (i) and multiply it by 2 to get the enthalpy change when 2 moles of NO2 are formed from 1 mole of N2 and 2 moles of O2:
2N2O2 (g) → 2N2 (g) + 4O2 (g) △H = +237.4 kJ

Now, we can sum up the enthalpy changes:
N2O4 (g) → 2NO2 (g) △H = +101.9 kJ
2N2O2 (g) → 2N2 (g) + 4O2 (g) △H = +237.4 kJ

By adding these two equations, we can cancel out the NO2 on both sides and get the overall equation for the formation of N2O4 from its elements:
N2O4 (g) + N2 (g) + 2O2 (g) △H = +339.3 kJ

Since we want to find the enthalpy of formation of N2O4, which is the enthalpy change when 1 mole of N2O4 is formed, we divide the above equation by 2:
1/2 N2O4 (g) + 1/2 N2 (g) + O2 (g) △H = +169.7 kJ

Therefore, the enthalpy of formation of N2O4 is +169.7 kJ.