Question

What is the enthalpy of formation (?H?f) of NaHCO3(s) from its constituent elements.

Answer 1

Answer:

Enthalpy of formation = -947.68KJ/mol

Explanation:

Enthalpy of formation is the heat change when one mole of a substance is formed from its element in its standard states and in standard conditions of temperature and pressure. it may be positive or negative, if positive, it is an endothermic reaction where the heat content of the product is greater than that of the reactants, and if negative, it is exothermic reaction - where the heat content of the reactants is greater than the products. the enthalpy of formation is measured in KiloJoule/Moles (KJ/Mole).

From the value of the enthalpy of formation of NaHCO3, it shows that the reaction is exothermic, that is the formation of NaHCO3 from its constituents elements. As such, the heat content of the reactants is greater than the products.

The step by step explanation is shown in the attachment.

Answer 2

Answer:

-947.6 kJ/mol.

Explanation:

The constituent elements combined to result in NaHCO3(s)

NaHCO3(s) → (1/2).Na2CO3(s) + (1/2).CO2(g) + (1/2).H2O(g) : ΔHr = +64.6 kJ/mol

Enthalpy change of a reaction can be measured and given the symbol,  . It is equal to the difference in enthalpy between reactants and products

From this equation we deduct that water is produced as a vapour, hence the reaction is endotermic and ΔHr is positive.

Therefore, To find the enthalpy of formation of NaHCO3 from this equation, we use the general equation which relates the enthalpy change ΔHr of the reaction to the enthalpies ΔHf of formation for the reactants and products.

ΔHr = ∑ΔH(products) - ∑ΔH(reagents)

where the  

ΔH is equal to the change in the potential energy of enthalpy of the chemical bond.

In ideal test scenarios, these values should be supplied, as you will not be able to look them up. Here, I've supplied values here.

ΔHf : Na2CO3(s) = -1130.7, CO2(g) = -393.5, H2O(g) = -241.8 kJ/mol

Using these figures (you may wish to check them in Wikipedia or elsewhere), we get

+64.6 = [(1/2)*(-1130.7) + (1/2)*(-393.5) + (1/2)*(-241.8)] - ΔHf(NaHCO3)

With the result that the enthalpy of formation of NaHCO3 is obtained as -947.6 kJ/mol.