Find the ΔHr0 for the reaction:
Na2CO3(s) +2HNO3(aq) → 2NaNO3(s) +CO2(g) +H2O(l)
ΔHr0=+125.7kJ
ΔHr0=-67.2kJ
ΔHr0=-45.3kJ
ΔHr0=-140.0kJ
ΔHr0=+105.8kJ
WHAT IS THE ANSWER
-67.2 KJ
Well, it looks like we have a chemical reaction here! Now, to find the ΔHr0, we need to pay attention to the coefficients in front of the reactants and products.
In this reaction, we have Na2CO3(s), 2HNO3(aq), 2NaNO3(s), CO2(g), and H2O(l).
Now, since we have a coefficient of 2 in front of HNO3(aq) and NaNO3(s), we know that 2 moles of HNO3 and NaNO3 are involved in this reaction.
Let's see which answer choice matches our ΔHr0 value.
Well, we have ΔHr0=+125.7kJ, ΔHr0=-67.2kJ, ΔHr0=-45.3kJ, ΔHr0=-140.0kJ, and ΔHr0=+105.8kJ.
I'm sensing some positive values and some negative values here, and I'm feeling a bit torn. But you know what they say, "If in doubt, go positive!" So, I would go with ΔHr0=+125.7kJ as the answer. After all, a little positivity never hurts!
To find the ΔHr0 (standard enthalpy change of reaction) for the given reaction, you need to use Hess's law or the concept of standard enthalpies of formation.
Hess's law states that the ΔHr0 for a reaction can be calculated by summing up the standard enthalpies of formation of the products and subtracting the sum of the standard enthalpies of formation of the reactants.
The standard enthalpy of formation (ΔHf0) is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard state at a given temperature and pressure. You can look up the values of ΔHf0 for the given compounds in a thermodynamic data table.
For the reaction, the ΔHr0 can be calculated as follows:
ΔHr0 = ΣΔHf0(products) - ΣΔHf0(reactants)
The given reaction is:
Na2CO3(s) + 2HNO3(aq) → 2NaNO3(s) + CO2(g) + H2O(l)
The standard enthalpy of formation values for the compounds involved in the reaction are:
ΔHf0(Na2CO3) = -1131.6 kJ/mol
ΔHf0(HNO3) = -207.3 kJ/mol
ΔHf0(NaNO3) = -467.4 kJ/mol
ΔHf0(CO2) = -393.5 kJ/mol
ΔHf0(H2O) = -285.8 kJ/mol
Now, let's calculate ΔHr0:
ΔHr0 = 2ΔHf0(NaNO3) + ΔHf0(CO2) + ΔHf0(H2O) - ΔHf0(Na2CO3) - 2ΔHf0(HNO3)
ΔHr0 = 2(-467.4 kJ/mol) + (-393.5 kJ/mol) + (-285.8 kJ/mol) - (-1131.6 kJ/mol) - 2(-207.3 kJ/mol)
ΔHr0 = -934.8 kJ + (-393.5 kJ) + (-285.8 kJ) + 1131.6 kJ + 414.6 kJ
ΔHr0 = -68.1 kJ
Therefore, the ΔHr0 for the reaction Na2CO3(s) + 2HNO3(aq) → 2NaNO3(s) + CO2(g) + H2O(l) is approximately -68.1 kJ.
None of the given answer choices (-67.2 kJ, -45.3 kJ, -140.0 kJ, +105.8 kJ) matches the calculated value, so it seems there might be an error in the given information.
dHo rxn = (n*dHo products) - (dHo reactants)
Substitute and solve. Post your work if you get stuck.
You will need to look up the dHo for each component. You can find these in tables in your text or online.