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?
I think it's -86, but I just want to make sure. Thanks in advance!
btw, bond energy is 389 for N-H, 436 FOR H-H, and 163 for N-N
Did you read the "be careful" part of the question? I think you have used N-N but the N2 molecules is NtriplebondN.
Using your numbers I end up with 80 kJ. I tried moving numbers around to obtain your value and could not duplicate it.
Also I looked up delta Hf for NH3 and it is about -46 kJ/mol or about +92 kJ for the reaction in reverse; i.e., the reaction shown. dHf values are more accurate so this is in the right ball park I think.
Well, you're not completely N2 it, but you're close! The ΔH for the reaction is actually -92 kJ. As for the kind of bond that nitrogen molecule makes, it forms a delightful and very stable triple bond. So you could say that it's a "nitro-genuine" bond! Keep up the chemistry curiosity, my friend!
To find the approximate ΔH (enthalpy change) for the given reaction, we first need to determine the enthalpy change for each individual bond broken and formed in the reaction.
The nitrogen molecule (N2) is known to form a triple bond. Breaking this triple bond requires energy, while forming new bonds between hydrogen and nitrogen atoms releases energy.
In the reaction, we have 3 hydrogen-hydrogen bonds (H-H) that are broken, requiring energy.
We also have 1 nitrogen-nitrogen triple bond (N≡N) that is broken.
On the other hand, we have 4 hydrogen-nitrogen bonds (H-N) that are formed, releasing energy.
The enthalpy change for each bond broken/formed is as follows:
Bond broken:
3 H-H bonds = 3 × ΔH(H-H)
Bond formed:
4 H-N bonds = 4 × ΔH(H-N)
1 N≡N bond = 1 × ΔH(N≡N)
To obtain the overall enthalpy change (ΔH) for the reaction, we subtract the sum of the energy released (bond formation) from the sum of the energy required (bond breaking), since bond breaking is an endothermic process and bond formation is an exothermic process.
Therefore, the approximate ΔH for the given reaction can be calculated as:
ΔH = (3 × ΔH(H-H)) - (4 × ΔH(H-N) + ΔH(N≡N))
To calculate the actual values for ΔH, we would need to refer to experimental data or thermodynamic tables. The value you provided, -86 kJ, might be an approximation or an actual value determined experimentally.
However, without the specific values for ΔH(H-H), ΔH(H-N), and ΔH(N≡N), it is not possible to confirm if -86 kJ is the correct value for the approximate ΔH in this reaction.
To determine the approximate ΔH for the given reaction, you need to calculate the difference in enthalpy between the products and the reactants.
First, you can look up the standard enthalpies of formation (ΔHf) for each compound involved in the reaction. The standard enthalpy of formation is the change in enthalpy when one mole of a substance is formed from its constituent elements in their standard states.
Using this, you can calculate the ΔH for the reaction using the following equation:
ΔH = Σ(ΔHf products) - Σ(ΔHf reactants)
Σ represents the summation. In this case, you'll sum up the standard enthalpies of formation for the products and subtract the sum of the standard enthalpies of formation for the reactants.
Now, let's calculate it step by step:
Reactants:
2NH3(g): Since ΔHf of NH3 is -46 kJ/mol (source: CRC Handbook of Chemistry and Physics), the ΔHf of 2NH3(g) is 2*(-46 kJ/mol).
Products:
3H2(g): The ΔHf of H2 is 0 kJ/mol (source: CRC Handbook of Chemistry and Physics), so the ΔHf of 3H2(g) is 3*(0 kJ/mol).
N2(g): The ΔHf of N2 is 0 kJ/mol (source: CRC Handbook of Chemistry and Physics), so the ΔHf of N2(g) is 0 kJ/mol.
Now, substitute these values into the equation:
ΔH = [Σ(ΔHf products)] - [Σ(ΔHf reactants)]
= [3*(0 kJ/mol) + 0 kJ/mol] - [2*(-46 kJ/mol)]
= 0 kJ/mol - (-92 kJ/mol)
= 92 kJ/mol
Therefore, the approximate ΔH for the given reaction is 92 kJ.
Regarding your second question about the bond of the nitrogen molecule (N2), it forms a triple covalent bond.