Nitric acid, HNO3, was first prepared 1200 years ago by heating naturally occurring sodium nitrate (called saltpeter) with sulfuric acid to produce sodium bisulfate and collecting the vapors of HNO3 produced. Calculate ΔH°rxn for this reaction. ΔH°f[NaNO3(s)] = -467.8 kJ/mol; ΔH°f[NaHSO4(s)] = -1125.5 kJ/mol; ΔH°f[H2SO4(l)] = -814.0 kJ/mol; ΔH°f[HNO3(g)] = -135.1 kJ/mol.
In all of my years I have never heard of NaNO3 being referred to as saltpeter. KNO3 yes but not NaNO3. One web site where I looked just now lists EITHER NaNO3 or KNO3 (or other nitrates); another lists just KNO3. Just a side note.
NaNO3 + H2SO4 ==> NaHSO4 + HNO3
dHrxn = (n*dHf products) - (n*dHf reactants)
To calculate ΔH°rxn for the reaction, we need to use the Hess's Law, which states that the enthalpy change for a reaction is the same regardless of the path taken to obtain the products.
The balanced chemical equation for the reaction is:
2 NaNO3(s) + H2SO4(l) -> Na2SO4(s) + 2 HNO3(g)
According to Hess's Law, the enthalpy change for the reaction can be calculated by considering the enthalpy changes for the formation of the reactants and products.
The enthalpy change for the reaction can be calculated using the following formula:
ΔH°rxn = ∑ΔH°f(products) - ∑ΔH°f(reactants)
where ∑ denotes the sum, ΔH°f denotes the standard enthalpy of formation.
Let's calculate the ΔH°rxn for this reaction:
∑ΔH°f(products) = [ΔH°f(Na2SO4(s))] + 2[ΔH°f(HNO3(g))]
∑ΔH°f(reactants) = 2[ΔH°f(NaNO3(s))] + [ΔH°f(H2SO4(l))]
Substituting the given values:
∑ΔH°f(products) = [0 kJ/mol] + 2[(-135.1 kJ/mol)]
∑ΔH°f(reactants) = 2[(-467.8 kJ/mol)] + [(-814.0 kJ/mol)]
Calculating the sums:
∑ΔH°f(products) = -270.2 kJ/mol
∑ΔH°f(reactants) = -1750.6 kJ/mol
ΔH°rxn = -270.2 kJ/mol - (-1750.6 kJ/mol)
ΔH°rxn = 1480.4 kJ/mol
Therefore, the ΔH°rxn for this reaction is 1480.4 kJ/mol.
To calculate ΔH°rxn for the reaction, we need to use the Hess's Law of heat summation. According to Hess's Law, the enthalpy change of a reaction can be calculated by taking the difference between the sum of the enthalpies of formation of the products and the sum of the enthalpies of formation of the reactants.
The balanced equation for the reaction is:
2 NaNO3(s) + H2SO4(l) -> Na2SO4(s) + 2 HNO3(g)
Now, let's calculate the enthalpy change using the enthalpies of formation provided:
Reactants:
ΔH°f[NaNO3(s)] = -467.8 kJ/mol
ΔH°f[H2SO4(l)] = -814.0 kJ/mol
Products:
ΔH°f[Na2SO4(s)] = unknown
ΔH°f[HNO3(g)] = -135.1 kJ/mol
Applying Hess's Law, we need to reverse the sign of the enthalpies of formation for the reactants and multiply them by their respective stoichiometric coefficients. For the products, we use the enthalpies of formation as is.
ΔH°rxn = (2 * ΔH°f[Na2SO4(s)]) + (2 * ΔH°f[HNO3(g)]) - (ΔH°f[NaNO3(s)]) - (ΔH°f[H2SO4(l)])
Now, let's substitute the given values into the equation:
ΔH°rxn = (2 * ΔH°f[Na2SO4(s)]) + (2 * -135.1 kJ/mol) - (-467.8 kJ/mol) - (-814.0 kJ/mol)
Simplifying further:
ΔH°rxn = (2 * ΔH°f[Na2SO4(s)]) - 135.1 kJ/mol + 467.8 kJ/mol - 814.0 kJ/mol
Now, we need to isolate the value of ΔH°f[Na2SO4(s)]. Rearranging the equation:
ΔH°rxn = (2 * ΔH°f[Na2SO4(s)]) - 481.3 kJ/mol
Finally, solving for ΔH°f[Na2SO4(s)]:
ΔH°f[Na2SO4(s)] = (ΔH°rxn + 481.3 kJ/mol) / 2
Note: The value of ΔH°f[Na2SO4(s)] is needed to calculate the enthalpy change in this reaction. As it is not provided, we cannot determine the exact value without further information.