Question

The enthalpy of formation of ${H^ + }(g)$ is $1536\,kJmo{l^{ - 1}}$. What is the heat of formation of ${H_2}(g)$ gas?
A. $3072\,kJ$
B. $1536\,kJ$
C. $0$
D. $- 1536\;kJ$
E. $- 3072\,kJ$

Answer 1

A change in the enthalpy for the formation of one mole of a compound from an element or a molecule present in its naturally existing state, is known as enthalpy of formation. It is represented by a symbol $\Delta {H_f}$. The sign convention for the enthalpy of formation is the same as that for change in enthalpy of the system.

Complete answer: When the enthalpy of formation is measured on formation of one mole of substance under standard conditions from its pure elements, then it is known as standard enthalpy of formation. It is represented by the symbol ${\Delta ^o}{H_f}$, where each notation has the following significance:
Delta $(\Delta )$ represents the change in enthalpy.
Degree ${(^o})$ signifies that the change is considered under standard conditions.
$f$ indicates that it is enthalpy of formation of a compound.
It is given that the enthalpy of formation of ${H^ + }(g)$ ion is $1536\,kJmo{l^{ - 1}}$ but the enthalpy of formation of ${H_2}$ gas will be zero because it is given in the phase in which hydrogen naturally exists in the atmosphere and we know that, the standard enthalpy of formation of pure elements or the molecules existing in their natural phase is considered to be zero.
Hence, the heat of formation of ${H_2}(g)$ gas $= 0$
So, option (C) is the correct answer.

Note:
It is important to note that the standard enthalpy change of reaction is equal to the difference between the summation of standard enthalpies of formation of products and reactants. The equation is expressed as ${\Delta ^o}{H_{reaction}} = \sum {{\Delta ^o}{H_f}{\text{(products)}}} - \sum {{\Delta ^o}{H_f}{\text{(reactants)}}}$.