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.

I answered this below earlier. If you don't understand something in my response you're welcome to ask for clarification but posting again just means you want someone to do the work for you.

difjsdkfjlskdf

To calculate ΔH°rxn for the reaction, we need to apply the Hess's Law, which states that the enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps (reactions) that make up the overall reaction.

The given reaction can be written as follows:
NaNO3(s) + H2SO4(l) -> NaHSO4(s) + HNO3(g)

Step 1: NaNO3(s) -> NaNO3(aq) (Dissolving NaNO3 in water)
ΔH1 = ΔHsol[NaNO3(s)] (Enthalpy of dissolution)

Step 2: NaNO3(aq) + H2SO4(l) -> NaHSO4(aq) + HNO3(aq) (In the aqueous phase)
ΔH2 = ΔHrxn

Step 3: NaHSO4(aq) -> NaHSO4(s) (Precipitation of NaHSO4)
ΔH3 = -ΔHsol[NaHSO4(s)] (Enthalpy of precipitation, opposite sign because it is the reverse of dissolution)

Step 4: HNO3(aq) -> HNO3(g) (Vaporizing HNO3)
ΔH4 = ΔHvap[HNO3(g)] (Enthalpy of vaporization)

Now, we can sum up the enthalpy changes of each step to calculate ΔH°rxn for the overall reaction:

ΔH°rxn = ΔH1 + ΔH2 + ΔH3 + ΔH4

Given enthalpy changes:
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

Let's substitute these values into the equation and calculate ΔH°rxn:

ΔH°rxn = ΔHsol[NaNO3(s)] + ΔHrxn + ΔHsol[NaHSO4(s)] + ΔHvap[HNO3(g)]

ΔH°rxn = 0 + ΔH2 + (-1125.5) + (-135.1)

Since the reaction involves the formation of NaNO3(s) and HNO3(g), we can assume that both NaNO3(s) and HNO3(g) have enthalpies of formation equal to zero. Therefore, the ΔHsol[NaNO3(s)] and ΔHvap[HNO3(g)] terms drop out.

ΔH°rxn = ΔH2 + (-1125.5) + (-135.1)

Substituting the given values:
ΔH°rxn = (-814.0) + (-1125.5) + (-135.1)
ΔH°rxn = -2074.6 kJ/mol

Therefore, the ΔH°rxn for the reaction is -2074.6 kJ/mol.

To calculate the standard enthalpy change (ΔH°rxn) for the reaction, you need to use the Hess's Law. Hess's Law states that the enthalpy change of a reaction is the same regardless of the route taken, as long as the initial and final conditions are the same.

The given reaction can be broken down into two steps:

Step 1: Formation of sodium bisulfate (NaHSO4) from sodium nitrate (NaNO3) and sulfuric acid (H2SO4):
NaNO3(s) + H2SO4(l) → NaHSO4(s)

Step 2: Formation of nitric acid (HNO3) from sodium bisulfate (NaHSO4):
NaHSO4(s) + HNO3(g) → H2SO4(l) + NaNO3(g)

Since the ΔH°f values are given for each species, you can use the ΔH°f values of the reactants and products to calculate the enthalpy change for each step.

Step 1:
ΔH°1 = ΔH°f[NaHSO4(s)] - ΔH°f[NaNO3(s)] - ΔH°f[H2SO4(l)]

Step 2:
ΔH°2 = ΔH°f[HNO3(g)] - ΔH°f[NaNO3(g)] - ΔH°f[H2SO4(l)]

Finally, you can calculate the overall enthalpy change (ΔH°rxn) by summing the enthalpy changes for both steps:

ΔH°rxn = ΔH°1 + ΔH°2

Let's calculate it using the given ΔH°f values:

Step 1:
ΔH°1 = (-1125.5 kJ/mol) - (-467.8 kJ/mol) - (-814.0 kJ/mol) = -174.7 kJ/mol

Step 2:
ΔH°2 = (-135.1 kJ/mol) - (-467.8 kJ/mol) - (-814.0 kJ/mol) = 212.9 kJ/mol

Overall enthalpy change:
ΔH°rxn = -174.7 kJ/mol + 212.9 kJ/mol = 38.2 kJ/mol

Therefore, the value of ΔH°rxn for the reaction is 38.2 kJ/mol.