Question:
For the reaction N$_2$ + 3H$_2$ $\rightarrow$ 2NH$_3$, $\Delta$H = ?
For the reaction N$_2$ + 3H$_2$ $\rightarrow$ 2NH$_3$, $\Delta$H = ?
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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, $\Delta H = \Delta E$.
Updated On: Jun 3, 2026
- $\Delta$E + 2RT
- $\Delta$E - 2RT
- $\Delta$H = RT
- $\Delta$E - RT
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The Correct Option is B
Solution and Explanation
Concept:
The relationship between enthalpy change ($\Delta H$) and internal energy change ($\Delta E$ or $\Delta U$) for a gaseous reaction is given by the formula:
\[ \Delta H = \Delta E + \Delta n_g RT \]
Where:
• $\Delta n_g$ is the change in the number of moles of gaseous products and reactants.
• $R$ is the universal gas constant.
• $T$ is the absolute temperature.
Step 1: Calculate $\Delta n_g$ for the given reaction.
The chemical equation is: \[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \]
• Moles of gaseous products ($n_p$) = $2$
• Moles of gaseous reactants ($n_r$) = $1 (\text{for N}_2) + 3 (\text{for H}_2) = 4$ \[ \Delta n_g = n_p - n_r = 2 - 4 = -2 \]
Step 2: Substitute $\Delta n_g$ into the relationship formula.
Using the formula $\Delta H = \Delta E + \Delta n_g RT$: \[ \Delta H = \Delta E + (-2)RT \] \[ \Delta H = \Delta E - 2RT \]
• $\Delta n_g$ is the change in the number of moles of gaseous products and reactants.
• $R$ is the universal gas constant.
• $T$ is the absolute temperature.
Step 1: Calculate $\Delta n_g$ for the given reaction.
The chemical equation is: \[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \]
• Moles of gaseous products ($n_p$) = $2$
• Moles of gaseous reactants ($n_r$) = $1 (\text{for N}_2) + 3 (\text{for H}_2) = 4$ \[ \Delta n_g = n_p - n_r = 2 - 4 = -2 \]
Step 2: Substitute $\Delta n_g$ into the relationship formula.
Using the formula $\Delta H = \Delta E + \Delta n_g RT$: \[ \Delta H = \Delta E + (-2)RT \] \[ \Delta H = \Delta E - 2RT \]
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