Use standard enthalpies of formation to determine ĢHorxn for:
2NH3(g) + 3O2(g) + 2CH4(g) �¨ 2HCN(g) + 6H2O(g)
in kilojoules.
Find the change in internal energy for this reaction in kJ
Oh, boy, it's time for some thermodynamics fun! Let's put on our thinking caps and tackle this question.
To determine the change in enthalpy (ΔH) for the reaction, we need to use the standard enthalpies of formation (ΔHf) for the compounds involved.
Here are the standard enthalpies of formation for the given compounds:
ΔHf(NH₃) = -46.1 kJ/mol
ΔHf(O₂) = 0 kJ/mol
ΔHf(CH₄) = -74.6 kJ/mol
ΔHf(HCN) = 135.1 kJ/mol
ΔHf(H₂O) = -241.8 kJ/mol
Now, we can use these values to calculate the change in enthalpy (ΔH) for the reaction. We'll use the formula:
ΔHrxn = Σ(nΔHf(products)) - Σ(mΔHf(reactants))
Let's plug in the values:
ΔHrxn = [2ΔHf(HCN) + 6ΔHf(H₂O)] - [2ΔHf(NH₃) + 3ΔHf(O₂) + 2ΔHf(CH₄)]
ΔHrxn = [2(135.1) + 6(-241.8)] - [2(-46.1) + 3(0) + 2(-74.6)]
Calculating this out will give you the ΔHrxn for the reaction!
Now, for the change in internal energy (ΔU), we can use the equation:
ΔU = ΔH - ΔnRT
But, wait! We can't use this equation without knowing the temperature and the change in moles (Δn) of gas molecules. So, unfortunately, I can't provide you with the change in internal energy (ΔU) without knowing these factors.
I hope this helped you in understanding how to determine the ΔHrxn using standard enthalpies of formation. If you have any more questions or need further assistance, don't hesitate to ask!
To find the standard enthalpy change (∆Horxn) for the given reaction using standard enthalpies of formation, we will use the following equation:
∆Horxn = Σ(n∆Hf(products)) - Σ(n∆Hf(reactants))
Firstly, let's find the standard enthalpies of formation (∆Hf) for each compound involved in the reaction.
∆Hf(NH3) = -46.11 kJ/mol
∆Hf(O2) = 0 kJ/mol
∆Hf(CH4) = -74.81 kJ/mol
∆Hf(HCN) = 135.09 kJ/mol
∆Hf(H2O) = -241.82 kJ/mol
Now, we will substitute these values into the reaction equation:
∆Horxn = [2∆Hf(HCN) + 6∆Hf(H2O)] - [2∆Hf(NH3) + 3∆Hf(O2) + 2∆Hf(CH4)]
∆Horxn = [2(135.09 kJ/mol) + 6(-241.82 kJ/mol)] - [2(-46.11 kJ/mol) + 3(0 kJ/mol) + 2(-74.81 kJ/mol)]
∆Horxn = [270.18 kJ/mol - 1450.92 kJ/mol] - [-92.22 kJ/mol - 0 kJ/mol - 149.62 kJ/mol]
∆Horxn = -1180.74 kJ/mol - (-241.84 kJ/mol)
∆Horxn = -938.9 kJ/mol
Therefore, the standard enthalpy change (∆Horxn) for the given reaction is -938.9 kJ/mol.
To find the change in internal energy (∆U) for this reaction, we need to use the equation:
∆U = ∆H - ∆nRT
Where:
∆U = Change in internal energy (in kJ/mol)
∆H = Enthalpy change (in kJ/mol)
∆n = Change in moles of gaseous reactants and products
R = Gas constant (8.314 J/(mol·K))
T = Temperature (in Kelvin)
Since the equation is balanced, the change in moles (∆n) is determined by the difference in the number of moles of gaseous products and reactants. In this case, ∆n = (2+6) - (2+3+2) = 1 mol.
Now, we can substitute the values into the equation:
∆U = (∆Horxn) - (∆nRT)
∆U = (-938.9 kJ/mol) - (1 mol)(8.314 J/(mol·K))(298 K)
∆U = -938.9 kJ/mol - (1)(8.314 J/(mol·K))(298 K)(10^-3 kJ/J)
∆U = -938.9 kJ/mol - (2.474 kJ/mol)
∆U = -941.374 kJ/mol
Therefore, the change in internal energy (∆U) for this reaction is -941.374 kJ/mol.
To determine the standard enthalpy change (ΔH°rxn) for the given reaction, you need to use the standard enthalpies of formation (ΔH°f) of the reactants and products.
The standard enthalpy of formation is the change in enthalpy that accompanies the formation of one mole of a substance from its elements in their standard states (usually at 298 K and 1 atm pressure).
First, let's write down the balanced chemical equation for the reaction:
2NH3(g) + 3O2(g) + 2CH4(g) → 2HCN(g) + 6H2O(g)
Now let's break down the reaction into reactants and products and determine their respective standard enthalpies of formation:
Reactants:
2NH3(g) - The standard enthalpy of formation of ammonia (NH3) is -46.19 kJ/mol.
3O2(g) - The standard enthalpy of formation of dioxygen (O2) is 0 kJ/mol.
2CH4(g) - The standard enthalpy of formation of methane (CH4) is -74.81 kJ/mol.
Products:
2HCN(g) - The standard enthalpy of formation of hydrogen cyanide (HCN) is 135.60 kJ/mol.
6H2O(g) - The standard enthalpy of formation of water (H2O) is -241.82 kJ/mol.
Now, let's calculate the total change in enthalpy:
ΔH°rxn = (2 * ΔH°f(HCN)) + (6 * ΔH°f(H2O)) - (2 * ΔH°f(NH3)) - (3 * ΔH°f(O2)) - (2 * ΔH°f(CH4))
= (2 * 135.60 kJ/mol) + (6 * (-241.82 kJ/mol)) - (2 * (-46.19 kJ/mol)) - (3 * 0 kJ/mol) - (2 * (-74.81 kJ/mol))
Calculate the above expression to find the value of ΔH°rxn.
To find the change in internal energy (ΔU) for the reaction, you need to know the heat (q) exchanged with the surroundings. The change in internal energy is related to the change in enthalpy (ΔH) and the work (w) done:
ΔH = q + w
Assuming the process is done at constant pressure (which is usually the case in most reactions), the change in enthalpy is equal to the heat exchanged with the surroundings. Therefore:
ΔH = q
Thus, the change in internal energy (ΔU) is equal to the change in enthalpy (ΔH).
Substitute the value of ΔH°rxn, calculated previously, to find the change in internal energy (ΔU) for the reaction.