The standard molar enthalpy of formation at 298 K is not zero for which of the following substances ?

Question

Accepted Answer

The standard molar enthalpy of formation is defined as the change in enthalpy when one mole of a substance is formed from its elements in their standard states at 298 K and 1 atm pressure. For elements in their most stable form at 298 K, the standard molar enthalpy of formation is zero.

1. **Option 1: $Br_{2}(g)$**  
   - **Explanation**: Bromine in its standard state at 298 K is a liquid, not a gas. Therefore, the formation of gaseous bromine from its standard state (liquid bromine) involves a phase change, which requires energy. This means that the standard molar enthalpy of formation for $Br_{2}(g)$ is not zero because it is not in its most stable form at 298 K. This justifies why Option 1 is correct.

2. **Option 2: C (Graphite)**  
   - **Explanation**: Graphite is the most stable form of carbon at 298 K. Therefore, the standard molar enthalpy of formation for graphite is zero because it is already in its standard state.

3. **Option 3: $Cl_{2}(g)$**  
   - **Explanation**: Chlorine gas ($Cl_{2}(g)$) is the most stable form of chlorine at 298 K. Thus, the standard molar enthalpy of formation for $Cl_{2}(g)$ is zero because it is in its standard state.

4. **Option 4: $Br_{2}(l)$**  
   - **Explanation**: Liquid bromine ($Br_{2}(l)$) is the standard state of bromine at 298 K. Therefore, the standard molar enthalpy of formation for $Br_{2}(l)$ is zero because it is in its most stable form.

In conclusion, Option 1 ($Br_{2}(g)$) is the correct answer because it is not in its standard state at 298 K, and thus its standard molar enthalpy of formation is not zero. The other options are in their standard states, and their standard molar enthalpies of formation are zero.