Determine the value of ∆H for the following reaction.

Question

Consider the generic reaction

                    2A+B → C+3D   ∆H_rxn= +120 kJ 

Determine the value of ∆H for the following reaction.

                    2C+6D → 4A+2B   ∆H_rxn = ?? kJ

Answer 1

Hello Ely!

Let's start with looking at the generic reaction. As you can see, we start with A and B and end up with C and D.

2A+B → C+3D   ∆H_rxn= +120 kJ 

However, in the other equation, we start with C and D and end up with A and B. 2C+6D → 4A+2B   ∆H_rxn = ?? kJ

To tackle this problem, we must first reverse the initial equation, getting:

C+3D → 2A + B

According to the laws of thermodynamics, when we reverse a chemical equation, we change the sign of its ∆H_rxn. Instead of ∆H_rxn= +120kJ, the reverse reaction therefore has ∆H_rxn= -120kJ. When looking at the second reaction in question, it seems all products and reactants have doubled in quantity from the generic reaction. To find the ∆H_rxn of the reaction, we simply double, so our final answer would be ∆H_rxn= -240kJ.

Hope this helps!