Select which of the following standard enthalpy of formation values is not zero at 25°C:

– Na(s)

– CH_4(g)

– H(g)

– Hg(l)

– Ne(g)

– S_8(rhombic)

This question aims to find out which of the above-mentioned elements/compounds has the standard enthalpy value not equal to zero at the given temperature. We will check if the given elements/compounds are in their standard state or not, after which we will define if their Standard Enthalpy of Formation is zero or not.

The Standard Enthalpy of Formation, also called the Standard Heat of Formation, is defined as the change in enthalpy when $1 M o l e$ of a substance in the standard state ( $1 a t m$ of pressure and $25^{\circ} C (298.15 K)$ ) is formed from its pure elements under the same conditions.

The symbol for the standard enthalpy of formation is $Δ H_{f}^{\circ}$

$Δ =$ A change in enthalpy

$^{\circ} =$ A degree that signifies it is a standard enthalpy change

$f =$ The f indicates that the substance is formed from its elements

Expert Answer:

The standard enthalpy of formation for a pure element in its reference form is zero as there is no chance involved in their formation. For example, the standard enthalpy of the formation of Carbon in graphite form is zero.

First, we will define the standard forms of the above-mentioned elements:

Sodium $N a$ – Standard form is $N a$ solid
Methane $C H_{4}$ comprises Carbon $C$ and Hydrogen $H$ . Carbon Standard form is $C$ whereas Hydrogen Standard form is $H_{2}$
Hydrogen $H$ – Standard form $H_{2}$ gaseous
Mercury $H g$ – Standard form $H g$ liquid
Neon $N e$ – Standard form $N e$ gaseous
Sulfur $S$ – Sulfur exists in rhombic standard state $S_{8}$

For the given elements, the following is their Standard Enthalpy of Formations.

a. $N a (s)$ – The standard form of sodium is in solid-state. Therefore, $Δ H_{f}^{\circ}$ $N a (s) = 0$ .
b. $C H_{4} (g)$ – It is not a pure element. Therefore, $Δ H_{f}^{\circ}$ $C H_{4}$ is not $0$ .
c. $H (g)$ – The standard form of hydrogen is $H_{2} (g)$ . Therefore, $Δ H_{f}^{\circ}$ $H (g)$ is not $0$ .
d. $H g (l)$ – The standard form of mercury is in liquid form. Therefore, $Δ H_{f}^{\circ}$ $H g (l) = 0$ .
e. $N e (g)$ – Neon exists in a gaseous standard state. Therefore, $Δ H_{f}^{\circ}$ $N e (g) = 0$ .
f. $S_{8} (r h o m b i c)$ – Sulfur exists in the rhombic standard state. Therefore, $Δ H_{f}^{\circ}$ $S_{8} (r h o m b i c) = 0$ .

Numerical Results

With this, the standard enthalpy of formation values for $C H_{4} (g)$ and $H (g)$ is not zero at $25^{\circ} C$ , as they are neither pure elements nor in their standard form.

Example:

Which of the following standard enthalpy of formation is not zero?

$C (g r a p h i t e)$
$H_{2} (g)$
$O_{2} (g)$
$H C L (g)$

Solution

The correct option is D – $H C l (g)$
As we now know, the standard enthalpy of formation for an element in its elemental state will always be $0$ .
Hence, $H C l (g)$ is not a pure element, so it will not have zero value of standard enthalpy of formation.

Expert Answer:

Numerical Results

Example:

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