Question

Show work, I need to learn how to do this and recognize the steps.

Calculate the free energy of formation of NaBr(s) given the following information: NaBr(s) → Na(s) + 1/2Br2(l), ΔG° = 349 kJ/mol

A) –309 kJ/mol

B) –329 kJ/mol

C) None of the above

D) –349 kJ/mol

E) –369 kJ/mol

Answer 1

Solution:

Given that,  

In the question it is said to find the free energy of formation of NaBr(s). So, we have to keep NaBr(s) in the product side and Na(s), Br₂(l) in the reactant side.

So, we have to reverse the reaction and change the sign on G.

The reaction will be-

From this reaction we can say that,  the free energy of formation of NaBr(s) is -349 kJ/mol.

Ans: (D) -349 kJ/mol

[Note: The standard Gibbs free energy of formation of a compound is the change of Gibbs free energy that accompanies the formation of 1 mole of a substance in its standard state from its constituent elements in their standard states at 1 bar of pressure and at a specified temperature.​​​​​​​]