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. The standard enthalpy of formation for an element in its most stable form is zero by definition.

1. **Option 3: $Br_{2}(g)$**  
   - **Explanation**: The standard state of bromine at 298 K is as a liquid, $Br_{2}(l)$. Therefore, the formation of gaseous bromine, $Br_{2}(g)$, from its standard state, $Br_{2}(l)$, involves a phase change from liquid to gas. This phase change requires energy, specifically the enthalpy of vaporization. As a result, the standard molar enthalpy of formation for $Br_{2}(g)$ is not zero because it is not in its standard state.

2. **Option 1: $Br_{2}(l)$**  
   - **Explanation**: Bromine in its standard state at 298 K is a liquid, $Br_{2}(l)$. Therefore, the standard molar enthalpy of formation for $Br_{2}(l)$ is zero because it is already in its most stable form at the given conditions.

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

4. **Option 4: $Cl_{2}(g)$**  
   - **Explanation**: Chlorine in its standard state at 298 K is a gas, $Cl_{2}(g)$. Therefore, the standard molar enthalpy of formation for $Cl_{2}(g)$ is zero because it is in its most stable form.

In conclusion, Option 3, $Br_{2}(g)$, 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.