2 HNO3(aq) + Na2CO3(s) 2 NaNO3(aq) + H2O(l) + CO2(g)

(a) Calculate H° for this reaction.
kJ

delta Hrxn = delta H products - delta H reactants.

i don't get how to find the delta H of the products or the reactants....do i have to figure them out or look them up?

To calculate the standard enthalpy change (ΔH°) for the given reaction, you need to use the enthalpy of formation values for the reactants and products.

The balanced equation for the reaction is:

2 HNO3(aq) + Na2CO3(s) → 2 NaNO3(aq) + H2O(l) + CO2(g)

The enthalpies of formation (ΔHf°) for the reactants and products can be found in tables, such as the NIST Chemistry WebBook. Here are the values:

ΔHf° HNO3(aq) = -207.2 kJ/mol
ΔHf° Na2CO3(s)= -1131.4 kJ/mol
ΔHf° NaNO3(aq) = -446.2 kJ/mol
ΔHf° H2O(l) = -285.8 kJ/mol
ΔHf° CO2(g) = -393.5 kJ/mol

Now, you can calculate the ΔH° for the reaction using the following equation:

ΔH° = Σ(nΔHf° products) - Σ(nΔHf° reactants)

where n is the stoichiometric coefficient for each reactant or product.

Applying this equation to the given reaction, we have:

ΔH° = (2 x ΔHf° NaNO3(aq)) + (1 x ΔHf° H2O(l)) + (1 x ΔHf° CO2(g)) - (2 x ΔHf° HNO3(aq)) - (1 x ΔHf° Na2CO3(s))

ΔH° = (2 x -446.2 kJ/mol) + (1 x -285.8 kJ/mol) + (1 x -393.5 kJ/mol) - (2 x -207.2 kJ/mol) - (1 x -1131.4 kJ/mol)

Calculating this expression will give you the value for ΔH° in kJ.

To calculate the standard enthalpy change (ΔH°) for a reaction, you need to use the enthalpy values of the reactants and products.

In this reaction, the balanced equation is:

2 HNO3(aq) + Na2CO3(s) → 2 NaNO3(aq) + H2O(l) + CO2(g)

H° represents the standard enthalpy change, or the enthalpy change that occurs when all reactants and products are in their standard states at a specified temperature (usually 25°C or 298 K).

To find the standard enthalpy change for this reaction, you'll need to use the enthalpy of formation (∆Hf°) values for each compound involved. The enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states.

Here are the ΔHf° values for the compounds involved in this reaction:

ΔHf° (HNO3(aq)) = -207.1 kJ/mol
ΔHf° (Na2CO3(s)) = -1130.9 kJ/mol
ΔHf° (NaNO3(aq)) = -467.7 kJ/mol
ΔHf° (H2O(l)) = -285.5 kJ/mol
ΔHf° (CO2(g)) = -393.5 kJ/mol

To calculate the ΔH° for the reaction, you need to consider the sum of the products and subtract the sum of the reactants:

ΔH° = [2 × ΔHf° (NaNO3(aq))] + [ΔHf° (H2O(l))] + [ΔHf° (CO2(g))] - [2 × ΔHf° (HNO3(aq))] - [ΔHf° (Na2CO3(s))]

Substituting the given values, you can calculate:

ΔH° = [2 × (-467.7 kJ/mol)] + (-285.5 kJ/mol) + (-393.5 kJ/mol) - [2 × (-207.1 kJ/mol)] - (-1130.9 kJ/mol)

After calculating the values and performing the arithmetic, you will find the standard enthalpy change (ΔH°) for this reaction.