Set up a Hess's law cycle, and use the following information to calculate ÄH°f for aqueous nitric acid, HNO3(aq). You will need to use fractional coefficients for some equations.
3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g) ÄH° = -138.4 kJ
2 NO(g) + O2(g) 2 NO2(g) ÄH° = -114 kJ
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(l) ÄH° = -1169.6 kJ
NH3(g) ÄH°f = -46.1 kJ/mol
H2O(l)
Where does the H2O(l) by itself at the bottom of your post fit in?
To set up a Hess's law cycle and calculate ΔH°f for aqueous nitric acid (HNO3), we need to manipulate the given chemical equations such that they add up to the formation of HNO3.
Here's how we can do it:
1. Start with the equation: 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g) ΔH° = -138.4 kJ
- Since we want to form HNO3, we'll keep this equation as it is.
2. Next, reverse equation 2: 2 NO(g) + O2(g) → 2 NO2(g) ΔH° = -114 kJ
- By reversing this equation, we can obtain NO2 as a reactant.
Reversed equation: 2 NO2(g) → 2 NO(g) + O2(g) ΔH° = +114 kJ
3. Multiply equation 2 by 2 and equation 1 by 2:
6 NO2(g) + 2 H2O(l) → 4 HNO3(aq) + 2 NO(g) ΔH° = -276.8 kJ (2 × -138.4 kJ)
4 NO(g) + 2 O2(g) → 4 NO2(g) ΔH° = -228 kJ (2 × -114 kJ)
4. Add equations 3 and 4 together:
6 NO2(g) + 2 H2O(l) + 4 NO(g) + 2 O2(g) → 4 HNO3(aq) + 2 NO(g) + 4 NO2(g) ΔH° = -276.8 kJ - 228 kJ
Simplifying the equation:
6 NO2(g) + 2 H2O(l) + 4 NO(g) + 2 O2(g) → 4 HNO3(aq) + 6 NO2(g) ΔH° = -504.8 kJ
5. We can cancel out the common species on both sides of the equation to obtain the formation of nitric acid:
2 H2O(l) + 4 NO(g) + 2 O2(g) → 4 HNO3(aq) ΔH° = -504.8 kJ
6. Finally, balance the equation by dividing all coefficients by 2:
H2O(l) + 2 NO(g) + O2(g) → 2 HNO3(aq) ΔH° = -252.4 kJ
Now, we can calculate ΔH°f for aqueous nitric acid (HNO3) using the given standard enthalpies of formation:
ΔH°f for NH3(g) = -46.1 kJ/mol
ΔH°f for H2O(l) = 0 kJ/mol (since it is not explicitly given)
Using the equation:
H2O(l) + 2 NO(g) + O2(g) → 2 HNO3(aq)
ΔH°f for HNO3(aq) = ΔH°(Products) - ΔH°(Reactants)
= [2 × ΔH°f(HNO3(aq))] - [ΔH°f(H2O(l)) + 2 × ΔH°f(NO(g)) + ΔH°f(O2(g))]
Substituting the values:
ΔH°f(HNO3(aq)) = [2 × ΔH°f(HNO3(aq))] - [0 + 2 × 0 + 2 × ΔH°f(NO(g)) + ΔH°f(O2(g))]
Now, we can't directly calculate ΔH°f for HNO3 since ΔH°f for NO and O2 are not given. However, if we are provided with ΔH°f values for NO and O2, we can substitute them into the equation and solve for ΔH°f for HNO3.
Note: In this case, without the given ΔH°f values for NO and O2, we cannot directly calculate ΔH°f for HNO3. Additional information is required to proceed with the calculation.
the h2o is
ÄH°f = -285.8 kJ/mol