Use the following information to answer the next 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
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.