Question

For the reaction N$_2$ + 3H$_2$ $\rightarrow$ 2NH$_3$, $\Delta$H = ?

• A. $\Delta$E + 2RT
• B. $\Delta$E - 2RT
• C. $\Delta$H = RT
• D. $\Delta$E - RT

Hint:

To find $\Delta n_g$ quickly, only sum the coefficients of substances in the gaseous state. If $\Delta n_g$ is negative, $\Delta H < \Delta E$; if positive, $\Delta H > \Delta E$; and if zero, $\Delta H = \Delta E$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Enthalpy (\(\Delta H\)) and Internal Energy (\(\Delta E\) or \(\Delta U\)) are related through the work done by pressure-volume changes.
For a reaction involving ideal gases, this relationship is expressed in terms of the change in the number of gaseous moles (\(\Delta n_g\)).
Step 2: Key Formula or Approach:
\[ \Delta H = \Delta E + \Delta n_g RT \]
Where \(\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})\).
Step 3: Detailed Explanation:
Given chemical equation:
\[ N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)} \]
Sum of gaseous coefficients of reactants \(= 1 (N_2) + 3 (H_2) = 4\).
Sum of gaseous coefficients of products \(= 2 (NH_3)\).
Calculate \(\Delta n_g\):
\[ \Delta n_g = 2 - 4 = -2 \]
Substitute into the relation:
\[ \Delta H = \Delta E + (-2)RT \]
\[ \Delta H = \Delta E - 2RT \]
Step 4: Final Answer:
The enthalpy change is \(\Delta E - 2RT\).