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:
$\Delta_{f}H^{\theta} = \frac{1}{2} \Delta_{r}H^{\theta}$
$\Delta_{f}H^{\theta} = \frac{1}{2} (-92.4 \, kJ \, mol^{-1})$
$\Delta_{f}H^{\theta} = -46.2 \, kJ \, mol^{-1}$
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Similar Questions
Find the bond enthalpy of $N-H$ bond in ammonia by using the change in enthalpy for the reaction given below: $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$; $\Delta H = -23 \ kcal$. Given bond energies: $N \equiv N = 226 \ kcal/mol$,$H-H = 103 \ kcal/mol$.
Medium
View SolutionAt $300 \, K$,the standard enthalpies of formation of $C_6H_5COOH_{(s)}$,$CO_{2_{(g)}}$,and $H_2O_{(l)}$ are $-408$,$-393$,and $-286 \, kJ \, mol^{-1}$ respectively. What is the heat of combustion of benzoic acid at constant volume in $kJ$ (in $.75$)? $(R = 8.31 \, J \, mol^{-1} \, K^{-1})$
Difficult
View SolutionCalculate the bond enthalpy of the $H-Cl$ bond from the following reaction:
$H_{2(g)} + Cl_{2(g)} \rightarrow 2 HCl_{(g)}$,$\Delta_{r} H^{\circ} = -185 \ kJ \ mol^{-1}$
(Given bond enthalpies of $H-H$ and $Cl-Cl$ bonds are $435.0 \ kJ \ mol^{-1}$ and $244 \ kJ \ mol^{-1}$ respectively.)
$H_{2(g)} + Cl_{2(g)} \rightarrow 2 HCl_{(g)}$,$\Delta_{r} H^{\circ} = -185 \ kJ \ mol^{-1}$
(Given bond enthalpies of $H-H$ and $Cl-Cl$ bonds are $435.0 \ kJ \ mol^{-1}$ and $244 \ kJ \ mol^{-1}$ respectively.)
The following two reactions are known:
$Fe_2O_{3(s)} + 3CO_{(g)} \rightarrow 2Fe_{(s)} + 3CO_{2(g)}; \Delta H = -26.8 \ kJ$
$FeO_{(s)} + CO_{(g)} \rightarrow Fe_{(s)} + CO_{2(g)}; \Delta H = -16.5 \ kJ$
The value of $\Delta H$ for the following reaction:
$Fe_2O_{3(s)} + CO_{(g)} \rightarrow 2FeO_{(s)} + CO_{2(g)}$ is ............. $kJ$.
$Fe_2O_{3(s)} + 3CO_{(g)} \rightarrow 2Fe_{(s)} + 3CO_{2(g)}; \Delta H = -26.8 \ kJ$
$FeO_{(s)} + CO_{(g)} \rightarrow Fe_{(s)} + CO_{2(g)}; \Delta H = -16.5 \ kJ$
The value of $\Delta H$ for the following reaction:
$Fe_2O_{3(s)} + CO_{(g)} \rightarrow 2FeO_{(s)} + CO_{2(g)}$ is ............. $kJ$.
MediumAIPMT 2010
View SolutionDiborane is formed from the elements as shown in equation $(i)$:
$2 B_{(s)} + 3 H_{2(g)} \longrightarrow B_2H_{6(g)} \dots (i)$
Given that:
$H_2O_{(l)} \longrightarrow H_2O_{(g)}, \quad \Delta H_1^{\circ} = 44 \, kJ$
$2 B_{(s)} + \frac{3}{2} O_{2(g)} \longrightarrow B_2O_{3(s)}, \quad \Delta H_2^{\circ} = -1273 \, kJ$
$B_2H_{6(g)} + 3 O_{2(g)} \longrightarrow B_2O_{3(s)} + 3 H_2O_{(g)}, \quad \Delta H_3^{\circ} = -2035 \, kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}, \quad \Delta H_4^{\circ} = -286 \, kJ$
The $\Delta H^{\circ}$ for the reaction $(i)$ is $..... \, kJ$.
$2 B_{(s)} + 3 H_{2(g)} \longrightarrow B_2H_{6(g)} \dots (i)$
Given that:
$H_2O_{(l)} \longrightarrow H_2O_{(g)}, \quad \Delta H_1^{\circ} = 44 \, kJ$
$2 B_{(s)} + \frac{3}{2} O_{2(g)} \longrightarrow B_2O_{3(s)}, \quad \Delta H_2^{\circ} = -1273 \, kJ$
$B_2H_{6(g)} + 3 O_{2(g)} \longrightarrow B_2O_{3(s)} + 3 H_2O_{(g)}, \quad \Delta H_3^{\circ} = -2035 \, kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}, \quad \Delta H_4^{\circ} = -286 \, kJ$
The $\Delta H^{\circ}$ for the reaction $(i)$ is $..... \, kJ$.
MediumKVPY 2010
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