
At 1 bar and ${\text{298 K}}$ , standard molar enthalpy of formation of which substance is zero?
A)${\text{H}}\left( {\text{g}} \right)$
B)${{\text{H}}^ + }\left( {{\text{aq}}} \right)$
C)${{\text{H}}^ + }\left( {\text{g}} \right)$
D)All correct
Answer
146.7k+ views
Hint:To solve this question, it is required to have knowledge about standard enthalpy of formation of any substance. The standard molar enthalpy of formation of a compound is defined as the enthalpy of formation of 1 mole of pure compound in its stable state at normal room temperature and pressure. At the normal temperature and pressure of the compound given, it is zero.
Complete step by step answer:
As we know that, 1 bar pressure and a temperature of ${\text{298 K}}$ are regarded as normal room temperature and pressure. So, we shall check each option for the correct answer:
In option A, the phase given is gaseous phase. We know that, in the gaseous phase at room temperature, hydrogen is present as ${{\text{H}}_2}$ . It is present as di-hydrogen has and not mono-hydrogen gas $\left( {\text{H}} \right)$. So, the molar enthalpy of formation is not zero.
In option B, the phase given is aqueous phase. We know that any polar compound containing hydrogen simultaneously dissociates into hydrogen ion and its conjugate base. Thus, ${{\text{H}}^ + }\left( {{\text{aq}}} \right)$ is normal at room temperature. So, the standard molar enthalpy of formation is zero.
In option C, the phase given is gaseous. We know that, in gaseous phase hydrogen exists as di-hydrogen gas and not in its ionic form. It is present as ${{\text{H}}_2}$ and not ${{\text{H}}^ + }$ . So, the standard molar enthalpy of formation is not zero.
Option D is not the correct option as well because option A and option C are incorrect.
$\therefore $ The correct option is option B, i.e. ${{\text{H}}^ + }\left( {{\text{aq}}} \right)$ at 1 bar and ${\text{298 K}}$ has a standard molar enthalpy of formation of zero.
Note: If the substance if present in a particular state at that phase, pressure and temperature, it is said that the standard molar enthalpy of formation of that compound is zero. For example, the molar enthalpy of formation of water in gaseous phase at 373 K will be zero. But for water in the liquid phase, the enthalpy of formation will be zero ay 298 K.
Complete step by step answer:
As we know that, 1 bar pressure and a temperature of ${\text{298 K}}$ are regarded as normal room temperature and pressure. So, we shall check each option for the correct answer:
In option A, the phase given is gaseous phase. We know that, in the gaseous phase at room temperature, hydrogen is present as ${{\text{H}}_2}$ . It is present as di-hydrogen has and not mono-hydrogen gas $\left( {\text{H}} \right)$. So, the molar enthalpy of formation is not zero.
In option B, the phase given is aqueous phase. We know that any polar compound containing hydrogen simultaneously dissociates into hydrogen ion and its conjugate base. Thus, ${{\text{H}}^ + }\left( {{\text{aq}}} \right)$ is normal at room temperature. So, the standard molar enthalpy of formation is zero.
In option C, the phase given is gaseous. We know that, in gaseous phase hydrogen exists as di-hydrogen gas and not in its ionic form. It is present as ${{\text{H}}_2}$ and not ${{\text{H}}^ + }$ . So, the standard molar enthalpy of formation is not zero.
Option D is not the correct option as well because option A and option C are incorrect.
$\therefore $ The correct option is option B, i.e. ${{\text{H}}^ + }\left( {{\text{aq}}} \right)$ at 1 bar and ${\text{298 K}}$ has a standard molar enthalpy of formation of zero.
Note: If the substance if present in a particular state at that phase, pressure and temperature, it is said that the standard molar enthalpy of formation of that compound is zero. For example, the molar enthalpy of formation of water in gaseous phase at 373 K will be zero. But for water in the liquid phase, the enthalpy of formation will be zero ay 298 K.
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