Question

The species which by definition has zero standard enthalpy of formation at 298K is :
(A) $B{{r}_{2}}(g)$
(B) $C{{l}_{2}}(g)$
(C) ${{H}_{2}}O(g)$
(D) $C{{H}_{4}}(g)$

Answer 1

Hint: The definition of standard state conditions specifies 1 atm of pressure, that liquids and gases be pure, and that solutions be at 1 M concentration. Temperature is not specified, although most tables compile data at 25 degrees C (298 K).

Complete step by step answer:
To know which of the following options have zero standard enthalpy of formation at 298K, we will consider each option one by one.
Considering option (A), $B{{r}_{2}}(g)$. Bromine is not gas at room temperature. It is in liquid form at 298K. Boiling point of Bromine is 331.8K, it will be gas above this temperature, therefore, its standard enthalpy of formation at 298K will not be zero.

Considering option (B), $C{{l}_{2}}(g)$. Chlorine is gas at room temperature. Boiling point of Chlorine is 238.96K. Therefore, its standard enthalpy of formation at 298K will be zero.

Considering option (C), ${{H}_{2}}O(g)$. Water is not gas at room temperature. It is in liquid form at 298K. Boiling point of Water is 373K, it will be gas above this temperature, therefore, its standard enthalpy of formation at 298K will not be zero.

Considering option (D), $C{{H}_{4}}(g)$. Methane is gas at room temperature. But its standard enthalpy of formation is in negative and not zero.
So, the correct answer is “Option B”.

Note: The standard enthalpy of formation is a measure of the energy released or consumed when one mole of a substance is created under standard conditions from its pure elements. STP stands for Standard Temperature and Pressure. It is defined to be 273 K (0 degrees Celsius) and 1 atm pressure (or 105 Pa).