Given
$N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)} \quad \Delta_{r}H^{\theta} = -92.4 \, kJ \, mol^{-1}$
What is the standard enthalpy of formation of $NH_{3}$ gas?
$N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)} \quad \Delta_{r}H^{\theta} = -92.4 \, kJ \, mol^{-1}$
What is the standard enthalpy of formation of $NH_{3}$ gas?
(C) The standard enthalpy of formation of a compound is the change in enthalpy that occurs during the formation of $1 \, mol$ of a substance in its standard state from its constituent elements in their standard states.
Rewriting the given equation for the formation of $1 \, mol$ of $NH_{3(g)}$:
$\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \longrightarrow NH_{3(g)}$
Therefore,the standard enthalpy of formation of $NH_{3(g)}$ is:
$\Delta_{f}H^{\theta} = \frac{1}{2} \Delta_{r}H^{\theta}$
$\Delta_{f}H^{\theta} = \frac{1}{2} (-92.4 \, kJ \, mol^{-1})$
$\Delta_{f}H^{\theta} = -46.2 \, kJ \, mol^{-1}$
Rewriting the given equation for the formation of $1 \, mol$ of $NH_{3(g)}$:
$\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \longrightarrow NH_{3(g)}$
Therefore,the standard enthalpy of formation of $NH_{3(g)}$ is:
$\Delta_{f}H^{\theta} = \frac{1}{2} \Delta_{r}H^{\theta}$
$\Delta_{f}H^{\theta} = \frac{1}{2} (-92.4 \, kJ \, mol^{-1})$
$\Delta_{f}H^{\theta} = -46.2 \, kJ \, mol^{-1}$
Explore More
Similar Questions
The heat of formation of $SiO_2$ and $MgO$ is respectively $-48.4 \ kJ$ and $-34.7 \ kJ$. The heat of the reaction $2Mg + SiO_2 \to 2MgO + Si$ is $...... \ kJ$.
Medium
View SolutionFrom the given reaction,$N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)} \quad \Delta H = -92.6 \ kJ$,the enthalpy of formation of $NH_3$ is (in $kJ$):
The enthalpy of neutralization of any strong acid and strong base is constant because......
Medium
View Solution$AB$,$A_2$,and $B_2$ are diatomic molecules. Enthalpies of dissociation of $AB$,$A_2$,and $B_2$ are in the ratio of $1:1:0.5$. Enthalpy of formation of $AB$,$\Delta_f H = -100 \ kJ \ mol^{-1}$. Find the dissociation enthalpy of $A_2$?
Reaction : $\frac{1}{2} A_2 + \frac{1}{2} B_2 \to AB$
Reaction : $\frac{1}{2} A_2 + \frac{1}{2} B_2 \to AB$
Medium
View SolutionCalculate the standard enthalpy change of the following reaction: $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(\ell)}$ if $\Delta_{f} H^{\circ}(CH_{4}) = -75 \ kJ \ mol^{-1}$,$\Delta_{f} H^{\circ}(CO_{2}) = -390 \ kJ \ mol^{-1}$,and $\Delta_{f} H^{\circ}(H_{2}O) = -286 \ kJ \ mol^{-1}$.
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo