1N2(g) reacts with H2(g) to form NH3(g), gaseous ammonia.
Show balanced thermochemical equations based on both one mole and two moles of NH3(g).
Determine Ho ( for both reactions using Table 6.2 (Hof).
Use the summation equations.
Show your calculations
N2 +H2 NH3 when I looked up the standard enthalpies of formation at 25 degrees the numbers came out to be 0. I don't know how to even set this up. Do I balance the equation?
Of course I don't know the contents of Table 6.2.
For 2 mols NH3 the equation is
N2 + 3H2 ==> 2NH3 (for 2 mols) dH = -46.11*2 kJ
1/2 N2 + 3/2 H2 ==> NH3 (for 1 mol) dH = -46.11 kJ
You should add (g) to each to show the phase.
I don't know how you got 0. My tables show dHof for NH3 = -46.11 kJ/mol and of course zero for H2 and N2.
I don't know what summation equations are looking at.
To balance the thermochemical equation for the reaction between N2(g) and H2(g) to form NH3(g), you need to make sure that the number of each element is balanced on both sides of the equation. The balanced equation for this reaction is:
N2(g) + 3H2(g) -> 2NH3(g)
Now, let's determine the ΔHo for this reaction using Table 6.2. The standard enthalpy of formation (ΔHof) values can be found in Table 6.2.
The ΔHof for N2(g) is 0 kJ/mol.
The ΔHof for H2(g) is 0 kJ/mol.
The ΔHof for NH3(g) is -46.0 kJ/mol.
To calculate the ΔHo for this reaction, you can use the formula:
ΔHo = ΣΔHof(products) - ΣΔHof(reactants)
For one mole of NH3(g):
ΔHo = (2ΔHof(NH3)) - (ΔHof(N2) + 3ΔHof(H2))
= (2(-46.0 kJ/mol)) - (0 kJ/mol + 3(0 kJ/mol))
= -92.0 kJ/mol
For two moles of NH3(g):
ΔHo = (2ΔHof(NH3)) - (ΔHof(N2) + 3ΔHof(H2))
= (2(-46.0 kJ/mol)) - (0 kJ/mol + 3(0 kJ/mol))
= -92.0 kJ/mol
So, the ΔHo for both reactions, with one mole and two moles of NH3(g), is -92.0 kJ/mol.
To determine the balanced thermochemical equations and calculate ΔHo for the reactions, we need to use the standard enthalpies of formation (ΔHof) values from Table 6.2. For the reaction of 1 mole of NH3(g), the balanced equation is:
N2(g) + 3H2(g) → 2NH3(g)
To find ΔHo for this reaction, we need to calculate the difference between the sum of the standard enthalpies of formation of the products and the sum of the standard enthalpies of formation of the reactants:
ΔHo = [2ΔHof(NH3) - ΔHof(N2) - 3ΔHof(H2)]
From Table 6.2, we find that ΔHof(NH3) = -45.9 kJ/mol, ΔHof(N2) = 0 kJ/mol, and ΔHof(H2) = 0 kJ/mol.
Substituting these values into the equation:
ΔHo = [2(-45.9 kJ/mol) - 0 kJ/mol - 3(0 kJ/mol)]
Simplifying:
ΔHo = -91.8 kJ/mol
Therefore, the ΔHo for the reaction of 1 mole of NH3(g) is -91.8 kJ/mol.
For the reaction of 2 moles of NH3(g), the balanced equation would be:
2N2(g) + 6H2(g) → 4NH3(g)
To find ΔHo for this reaction, we use the same formula:
ΔHo = [4ΔHof(NH3) - 2ΔHof(N2) - 6ΔHof(H2)]
Substituting the values:
ΔHo = [4(-45.9 kJ/mol) - 2(0 kJ/mol) - 6(0 kJ/mol)]
Simplifying:
ΔHo = -183.6 kJ/mol
Therefore, the ΔHo for the reaction of 2 moles of NH3(g) is -183.6 kJ/mol.
In both cases, it is not necessary to balance the equation. The standard enthalpies of formation provide the information needed to calculate ΔHo.