Question

(5) (1 point) Write the heat of formation reaction for disodium edetate, Na2C10H14N208.

Answer 1

The structure of C₁₀H₁₄N₂Na₂O₈ is:

credits: PubChem

N Na+ N H Na+ H-O

The heat of formation reaction for the said compound is:

Disodium EDTA is prepared by dissolving EDTA into a hot solution that contained two equivalents of sodium hydroxide. The solution was then allowed to crystallize.

EDTA+NaOH ---------->  Disodium EDTA + 2H₂O

C₁₀H₁₆N₂O₈ + 2NaOH --------> C₁₀H₁₄N₂Na₂O₈ +2H₂O

For the heat of formation: we need to find out the Heat of formation of EDTA and NaOH and H₂O

then for the answer:

Heat of formation is difference of heat between products and reactants

del(Disodium edetate)= del(edta)+ 2*del(NaOH) - 2*del(H₂O)

where del(X)= heat of formation of compound X