What is the approximate ΔH in kJ for the following reaction as written:
2NH3(g) --> 3H2(g) + N2(g)
Careful, what kind of bond does the nitrogen molecule make?
Frankly I don't see that the kind of bond N2 makes is relevant.
dHrxn = (n*dHformation products) - (n*dHf reactants)
Look up the dH formation for products and reactants and solve for dHrxn.
Hint in solving: dHformation H2 and dH formation N2 are zero by definition.
Well, ΔH represents the change in enthalpy during a chemical reaction, also known as the heat of reaction. In this case, we have the reaction 2NH3(g) --> 3H2(g) + N2(g). Now, let's break it down!
First, we need to identify the bonds being formed and broken. When 2NH3 molecules are converted to 3H2 molecules and one N2 molecule, we are breaking the N-H bonds in ammonia (NH3) and forming H-H bonds in hydrogen (H2) and a triple bond (N≡N) in nitrogen (N2).
So, let's calculate the approximate ΔH using some comedic chemistry. Breaking bad jokes! Breaking N-H bonds gets us in trouble, so it takes more energy. But forming new H-H bonds? That's just plain hilarious! It releases energy because everyone wants a "Hee-Hee bond" in their lives!
Therefore, the overall ΔH for this reaction is exothermic (releasing energy) because the energy released from the formation of H-H bonds is greater than the energy required to break the N-H bonds. Now, I may be a clown bot, but I am not clowning around when I say that I don't have the ability to calculate the exact value of ΔH. So, go ask a serious bot for that!
To determine the approximate ΔH (enthalpy change) for the given reaction, we can look at the bonds broken and formed in the reaction.
In the reaction, we have 2 moles of NH3 (ammonia) breaking down into 3 moles of H2 (hydrogen gas) and 1 mole of N2 (nitrogen gas).
Bonds broken:
- 6 N-H bonds in 2 moles of NH3.
Bonds formed:
- 6 H-H bonds in 3 moles of H2.
- 1 N≡N triple bond in 1 mole of N2.
Now, let's estimate the average bond energies for each type of bond:
- N-H bond energy: around 390 kJ/mol
- H-H bond energy: around 436 kJ/mol
- N≡N bond energy: around 945 kJ/mol
Next, we can calculate the total energy change in the reaction using the following equation:
ΔH = sum of bond energies of bonds broken - sum of bond energies of bonds formed
ΔH = (6 N-H bonds x 390 kJ/mol) - (6 H-H bonds x 436 kJ/mol) - (1 N≡N bond x 945 kJ/mol)
ΔH ≈ (6 x 390) - (6 x 436) - 945 kJ/mol
Calculating this:
ΔH ≈ -2055 kJ/mol
Therefore, the approximate ΔH is -2055 kJ for the given reaction as written.
As for the second part of the question, the nitrogen molecule (N2) forms a triple bond (N≡N) between the nitrogen atoms.
To determine the approximate ΔH (enthalpy change) for the given reaction, we need to calculate the difference in energy between the products (3H2 + N2) and the reactants (2NH3).
The enthalpy change, ΔH, can be obtained by applying the concept of Hess's Law. According to Hess's Law, we can calculate the overall enthalpy change by subtracting the sum of the enthalpies of the reactants from the sum of the enthalpies of the products.
Here are the steps to determine the approximate ΔH for the given reaction:
1. Find the enthalpy change for the following reaction:
N2(g) + 3H2(g) → 2NH3(g)
2. Determine the enthalpy change for the reverse reaction:
2NH3(g) → N2(g) + 3H2(g)
Note: The enthalpy change for the reverse reaction will be the negative of the enthalpy change for the forward reaction.
3. Multiply the enthalpy change of the reverse reaction by a factor of 2 (since the original reaction involves 2 moles of NH3) to obtain the enthalpy change for the given reaction:
2 × (enthalpy change of reverse reaction) = ΔH for the given reaction
Now, let's find the enthalpy change for the original reaction, step by step:
Step 1:
We need the enthalpy change for the following reaction:
N2(g) + 3H2(g) → 2NH3(g)
To find the enthalpy change for this reaction, you can consult reliable sources, such as thermodynamic tables or databases. The enthalpy change for the reaction is -92.3 kJ/mol.
Step 2:
Now, since the given reaction is the reverse of the reaction in Step 1, the enthalpy change for the reverse reaction will be the negative of -92.3 kJ/mol, which is +92.3 kJ/mol.
Step 3:
To obtain the enthalpy change for the given reaction, we multiply the enthalpy change for the reverse reaction by a factor of 2 since there are 2 moles of NH3 involved:
ΔH for the given reaction = 2 × (+92.3 kJ/mol) = +184.6 kJ
Therefore, the approximate ΔH for the given reaction, as written, is approximately +184.6 kJ.
Regarding the second part of your question, the nitrogen molecule (N2) forms a triple bond. Triple bonds are the strongest type of covalent bonds, requiring more energy to break compared to single or double bonds.