(A) The enthalpy of reaction is calculated using the formula: $\Delta H = \sum BE_{\text{reactants}} - \sum BE_{\text{products}}$.For the reaction $N_

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(A) The enthalpy of reaction is calculated using the formula: $\Delta H = \sum BE_{\text{reactants}} - \sum BE_{\text{products}}$.For the reaction $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$,the bonds broken are $1$ mole of $N \equiv N$ and $3$ moles of $H - H$,and the bonds formed are $6$ moles of $N - H$.$\Delta H = [BE(N \equiv N) + 3 \times BE(H - H)] - [6 \times BE(N - H)]$.Substituting the given values: $\Delta H = [945 + 3(436)] - [6(391)]$.$\Delta H = [945 + 1308] - [2346]$.$\Delta H = 2253 - 2346 = -93 \ kJ$.

Class 11 Chemistry Thermodynamics Heat of reaction, Bond energy and Hess law

Medium

Given the bond energies of $N \equiv N$,$H - H$ and $N - H$ bonds are $945$,$436$ and $391 \ kJ \cdot mol^{-1}$ respectively,the enthalpy of the following reaction $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$ is ...... $kJ$.

A
$-93$
B
$102$
C
$90$
D
$105$

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Class 11

Chemistry Questions

Chapter

Thermodynamics

Subtopic

Heat of reaction, Bond energy and Hess law

Equal volumes of $1 \, M \, HCl$ and $1 \, M \, H_2SO_4$ are neutralized by a dilute $NaOH$ solution,and $x$ and $y \, kcal$ of heat are liberated,respectively. Which of the following is correct?

Medium

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Identify the law for the statement "Overall the enthalpy change for a reaction is equal to the sum of enthalpy changes of individual steps in the reaction".

EasyMHT CET 2026

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The bond energies of $H-H$ and $Cl-Cl$ are $430 \, kJ/mol$ and $240 \, kJ/mol$ respectively. If $\Delta_f H$ for $HCl$ is $-90 \, kJ/mol$,then the bond enthalpy of $HCl$ will be ...... $kJ/mol$.

Medium

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Find the enthalpy of formation of the $OH^-$ ion in $KJ$ at $25^\circ C$ from the following data:
$H_2O_{(l)} \to H^+_{(aq)} + OH^-_{(aq)} ; \Delta H = 57.32 \ KJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(l)} ; \Delta H = -286.20 \ KJ$

Difficult

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At $25^{\circ} C$ and $1 \ atm$ pressure,the enthalpies of combustion are as given below:

Substance $H_{2(g)}$ $C_{\text{(graphite)}}$ $C_{2}H_{6(g)}$

$\Delta_{c}H^{\Theta} / (kJ \ mol^{-1})$ $-286.0$ $-394.0$ $-1560.0$

The enthalpy of formation of ethane is ........ .

MediumJEE MAIN 2022

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Substance	$H_{2(g)}$	$C_{\text{(graphite)}}$	$C_{2}H_{6(g)}$
$\Delta_{c}H^{\Theta} / (kJ \ mol^{-1})$	$-286.0$	$-394.0$	$-1560.0$

Given the bond energies of $N \equiv N$,$H - H$ and $N - H$ bonds are $945$,$436$ and $391 \ kJ \cdot mol^{-1}$ respectively,the enthalpy of the following reaction $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$ is ...... $kJ$.

Similar Questions

Equal volumes of $1 \, M \, HCl$ and $1 \, M \, H_2SO_4$ are neutralized by a dilute $NaOH$ solution,and $x$ and $y \, kcal$ of heat are liberated,respectively. Which of the following is correct?

Identify the law for the statement "Overall the enthalpy change for a reaction is equal to the sum of enthalpy changes of individual steps in the reaction".

The bond energies of $H-H$ and $Cl-Cl$ are $430 \, kJ/mol$ and $240 \, kJ/mol$ respectively. If $\Delta_f H$ for $HCl$ is $-90 \, kJ/mol$,then the bond enthalpy of $HCl$ will be ...... $kJ/mol$.

Find the enthalpy of formation of the $OH^-$ ion in $KJ$ at $25^\circ C$ from the following data:$H_2O_{(l)} \to H^+_{(aq)} + OH^-_{(aq)} ; \Delta H = 57.32 \ KJ$$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(l)} ; \Delta H = -286.20 \ KJ$

At $25^{\circ} C$ and $1 \ atm$ pressure,the enthalpies of combustion are as given below: Substance $H_{2(g)}$ $C_{\text{(graphite)}}$ $C_{2}H_{6(g)}$ $\Delta_{c}H^{\Theta} / (kJ \ mol^{-1})$ $-286.0$ $-394.0$ $-1560.0$ The enthalpy of formation of ethane is ........ .

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Find the enthalpy of formation of the $OH^-$ ion in $KJ$ at $25^\circ C$ from the following data:
$H_2O_{(l)} \to H^+_{(aq)} + OH^-_{(aq)} ; \Delta H = 57.32 \ KJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(l)} ; \Delta H = -286.20 \ KJ$

At $25^{\circ} C$ and $1 \ atm$ pressure,the enthalpies of combustion are as given below:

Substance $H_{2(g)}$ $C_{\text{(graphite)}}$ $C_{2}H_{6(g)}$

$\Delta_{c}H^{\Theta} / (kJ \ mol^{-1})$ $-286.0$ $-394.0$ $-1560.0$

The enthalpy of formation of ethane is ........ .