Use standard enthalpies of formation to determine ΔHorxnfor:
3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g)
-133.5 = Delta H
Find the change in internal energy for this reactioion.. ???? In KJ
Well, before we dive into the calculation, let me tell you a chemistry joke: Why did the noble gas cry? Because all its friends argon!
Now, let's determine the change in internal energy for this reaction. The change in internal energy (ΔU) can be calculated using the equation:
ΔU = Σ(nΔHf(products)) - Σ(nΔHf(reactants))
Where Σ(nΔHf(products)) is the sum of the standard enthalpies of formation of the products multiplied by their respective coefficients, and Σ(nΔHf(reactants)) is the sum of the standard enthalpies of formation of the reactants multiplied by their respective coefficients.
From the reaction equation provided, we have:
Reactants:
3NO2(g) + H2O(l)
Products:
2HNO3(aq) + NO(g)
Using the given ΔH value of -133.5 kJ, the standard enthalpies of formation can be found in a reference book or online resource.
Let's assume:
ΔHf(HNO3(aq)) = -207 kJ/mol
ΔHf(NO(g)) = 90 kJ/mol
ΔHf(NO2(g)) = 33 kJ/mol
ΔHf(H2O(l)) = -286 kJ/mol
Now, substitute the values into the equation:
ΔU = [2(-207 kJ/mol) + 90 kJ/mol] - [3(33 kJ/mol) + (-286 kJ/mol)]
After simplifying the equation and performing the arithmetic, you'll get the change in internal energy (ΔU) in kJ.
I hope that helps! Let me know if you have any more questions or if you'd like another joke.
To determine the change in internal energy (ΔU) for this reaction, you can use the equation:
ΔU = ΔH - ΔnRT
Where:
ΔH = enthalpy change of the reaction
Δn = change in the number of moles of gas
R = ideal gas constant (8.314 J/(mol·K))
T = temperature in Kelvin
Given that the enthalpy change (ΔH) for the reaction is -133.5 kJ, we can proceed to calculate the change in internal energy (ΔU).
First, we need to determine the change in the number of moles of gas (Δn). Looking at the balanced chemical equation:
3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g)
We can see that the number of moles of gas changes from 4 moles (3 moles of NO2 + 1 mole of H2O) on the reactant side to 3 moles (2 moles of HNO3 + 1 mole of NO) on the product side. Therefore, Δn = 3 - 4 = -1.
Assuming the temperature is constant, we can use the ideal gas constant R in units of kJ/(mol·K) to calculate ΔU.
ΔU = -133.5 kJ - (-1 mol) * (8.314 kJ/(mol·K)) * (298 K)
ΔU = -133.5 kJ + 2.475 kJ
ΔU ≈ -131.025 kJ
Therefore, the change in internal energy (ΔU) for the given reaction is approximately -131.025 kJ.
To find the change in internal energy (ΔU) for a reaction, we can use the equation:
ΔU = ΔH - ΔnRT
Where:
ΔH is the change in enthalpy of the reaction.
Δn is the change in the number of moles of gas.
R is the ideal gas constant (8.314 J/(mol·K)).
T is the temperature in Kelvin.
In this case, we already have the change in enthalpy (ΔH) of the reaction, which is given as -133.5 kJ.
Now we need to determine the change in the number of moles of gas (Δn) for this reaction. We do this by comparing the stoichiometric coefficients of the reaction.
From the balanced equation:
3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g)
We see that there is a decrease of 3 moles of NO2 gas (reactant), an increase of 2 moles of HNO3 gas (product), and an increase of 1 mole of NO gas (product).
Therefore, Δn = (Δmoles of gas products) - (Δmoles of gas reactants)
Δn = (2 + 1) - 3 = 0
Since Δn is 0 in this case, the second term (ΔnRT) in the equation becomes zero, and we are left with:
ΔU = ΔH - ΔnRT
ΔU = -133.5 kJ - 0
So, the change in internal energy (ΔU) for this reaction is also -133.5 kJ.
dHrxn = (n*dHreactants) - (n*dHproducts)
dE = dH + work
work is pdV but I don't see any pressure. dV you can get from 3 mols on the left going to 1 mol on the right.