Question

Consider the hypothetical thermochemical equation 3 A + B → 2 C for which ΔH = 65.5 kJ/mol.What would ∆H, in kJ/mol, be for the reaction 2 C → 3 A + B?

Answer 1

`-65.5\text{kJ/mol}`

Explanations:

Given the hypothetical thermochemical equation, expressed as:

`3A+B\rightarrow2C;\triangle H=65.5\text{kJ/mol}`

We are to find the enthalpy change (ΔH ) for the reaction 2C → 3 A + B.

• You can see that the, given reaction, 2C → 3 A + B is the, reverse ,of the thermochemical equation 3 A + B → 2 C.

• Since the ,reverse reaction is possible,, the reaction enthalpy has the ,same numerical value, but with the, ,opposite sign

Therefore, the enthalpy change ∆H, in kJ/mol, be for the reaction 2 C → 3 A + B is -65.5kJ/mol