Enthalpy - Chemistry

Question

Consider the following balanced reaction:

2 A₂ + 3 B → C₂ + 2 D

What is the enthalpy of reaction if the enthalpy of formation of the compounds is given as

Compound	Enthalpy of Formation (kJ/mol)
A2	6.61
B	-13.48
C2	0.00
D	22.45

Answer 1

Hi Julia! Remember enthalpy is a state function, so we can take as many paths as we want between the initial state and the final state, the value is going to be the same.

Let us take the path of assembling each compound. Assembling a compound can be thought of as forming a compound, so we can use the enthalpies of formation of each species and add/subtract them!

The final state of this system is the products, and the initial state is the reactants. Therefore, the delta Hrxn can be found:

delta Hrxn = (sum of delta Hf,i * vi) of the products - (sum of delta Hf,i * vi) of the reactants

where vi is the stoichiometric number of the chemical reaction. Therefore:

delta Hrxn = (1C2,f + 2D,f) - (2A2,f + 3B,f) = (2*22.45 kJ/mol + 1*0.00 kJ/mol) -(2*6.61 kJ/mol + 3* -13.48 kJ/mol)

= 44.90 kJ - (-27.22 kJ) = 72.12 kJ

I hope this helps!