Tag: bond enthalpies

Questions Related to bond enthalpies

If BE (bond energy) of N $\equiv $ N bond is ${ X } _{ 1 }$ that of H H bond is  ${ X } _{ 2 }$ and N H bond is ${ X } _{ 1 }$ then enthalpy change of the reaction is
${ N } _{ 2 }+{ 3H } _{ 2 }\longrightarrow { 2NH } _{ 3 }$
${ \Delta H } _{ r }={ X } _{ 1 }+{ 3X } _{ 2 }-{ 2X } _{ 3 }$

chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

  1. True

  2. False

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Correct Option: B
Explanation:

As we know,

Heat of reaction = bond energy of reactant - bond energy of product
${ N } _{ 2 }+{ 3H } _{ 2 }\longrightarrow { 2NH } _{ 3 }$
so
${ \Delta H } _{ r }={ X } _{ 1 }+{ 3X } _{ 2 }-{ 2\times3X } _{ 3 }$
${ \Delta H } _{ r }={ X } _{ 1 }+{ 3X } _{ 2 }-{ 6X } _{ 3 }$

The bond energy is the energy required to:

chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

  1. dissociation one liter of substance

  2. dissociate bond of 1 kg of substance

  3. break one mole of covalently bonded gas molecules

  4. break bonds in one gram of substance

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Correct Option: C
Explanation:

Bond energy is the measure of bond strength in a chemical bond. It is the energy or heat required to break one mole of molecules into their individual atoms.

The enthalpy change for the following reaction is 514 kJ. Calculate the average Cl - F bond energy. 


   $ClF _3(g)\rightarrow Cl(g)+3:F(g)$

chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

  1. 1542

  2. 88

  3. 171

  4. 514

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Correct Option: C
Explanation:
   $ClF _3(g)\rightarrow Cl(g)+3\:F(g)$

As there are three $Cl-F$ bond,

Therefore, average Cl - F bond energy $= \dfrac{514}{3} = 171$


Hence, option C is correct.

$AB,\, A _2$ and $B _2$ are diatomic molecules. If the bond enthalpies of $A _2,\, AB\, &\, B _2$ are in the ratio 1 : 1 : 0.5 and enthalpy of formation of AB from $A _2$ and $B _2$ is - 100 kJ/mol$^{-1}$. What is the bond enthalpy of $A _2$.

chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

  1. 400 kJ/mole

  2. 300 kJ/mole

  3. 500 kJ/mole

  4. 900 kJ/mole

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Correct Option: A
Explanation:

$\frac {1}{2} A _2 + \frac {1}{2}B _2 \rightarrow AB \Delta H= -100J$

Let the bond enthalpies of $A _2 : B _2 : AB = 1: 0.5 : 1 $
From reaction : $0.5x + 0.25x - 1x = 100 \Rightarrow x= \frac {100}{0.25}= 400kJ $
Bond enthalpy of $A _2= 1x = 400kJmol^{-1}$

From the following thermochemical equations:
$C _2H _4 \rightarrow C _2H _6;     \triangle H = -32.7 kcal$
$C _6H _6+3H _2 \rightarrow C _6H _{12};   \triangle H= -49.2kcal$
Calculate resonance energy of benzene.

chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

  1. $-48.9\ kcal$

  2. $-49.2\ kcal$

  3. $-16.5\ kcal$

  4. $-98.1\ kcal$

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Correct Option: A
Explanation:

Enthalpy of hydrogenation of $C=C$ bond $=-32.7\ kcal$

Calculated  enthalpy of hydrogenation of benzene $=-32.7\times3=-98.1\ kcal$

Actual enthalpy of hydrogenation of benzene$=-45.2\ kcal$
Hence, Resonance energy of benzene$=-98.1-(-45.2)=52.9\ kcal$

Energy required to dissociate $4g$ of gaseous hydrogen into free gaseous atoms is $208\ Kcal$ at ${25}^{o}C$. The bond energy of $H-H$ bond will be:

chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

  1. $1.04\ Kcal$

  2. $10.4\ Kcal$`

  3. $104\ Kcal$

  4. $1040\ Kcal$

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Correct Option: C
Explanation:

The reaction involved is,
$H _2(g) \rightarrow 2H(g)$                


  $\Delta H = \text{H-H bond energy}$

Since, $\text{moles =} \dfrac{\text{mass of compound}}{\text{molecular mass of compound}}$

So, $ \text{moles of} \ H _2 = \dfrac{4}{2} = 2$

Hence, energy required to break 4gm or 2 moles of $H _2$ into gaseous atoms = $208Kcal = 2\times \text{H-H bond energy}$

$\text {H-H bond energy}$ = $104Kcal $ 

Hence, answer is option C.

The dissociation energy of ${CH} _{4}$ is $400kcal$ ${mol}^{-1}$ and that of ethane is $670kcal$ ${mol}^{-1}$. The C-C bond energy is:

chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

  1. $270kcal$

  2. $70kcal$

  3. $200kcal$

  4. $240kcal$

Show answer
Correct Option: B
Explanation:

Dissociation reaction of $CH _4$ is,
$CH _4(g) \rightarrow C(g) + 4H(g)$         =>    $\Delta H = 400\text{ Kcal mol}^{-1} = 4\times \text{(bond energy of C-H bond)}$
=> $\text{bond energy of C-H bond = 100Kcal}$
Similarly, dissociation reaction of $C _2H _6$ is,
$C _2H _6(g) \rightarrow 2C(g) +6H(g)$     =>    $\Delta H = 670\text{Kcal mol}^{-1} =\text{(C-C bond energy)} + 6\times \text{(C-H bond energy)} $
=> $\text{(C-C bond energy)} = 670 - 6\times 100 = 70Kcal$ 
Hence, answer is option B.

The bond dissociation energy of $C-H$ in ${CH} _{4}$ from the equation
$C(g)+4H(g)\rightarrow {CH} _{4}(g);\Delta H=-397.8kcal$ is:

chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

  1. $+99.45kcal$

  2. $-99.45kcal$

  3. $+397.8kcal$

  4. $+198.9kcal$

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Correct Option: A
Explanation:

$\Delta H $ of the given reaction,
$C(g) + 4H(g) \rightarrow CH _4(g)$ 
can be written as, $\Delta H = \text {heat released in formation of 4 C-H bonds in CH} _4$
=> $\Delta H = -4\times \text{bond dissociation energy of C-H bond in CH} _4$
=> $-397.8 = -4\times \text{bond dissociation energy of C-H bond in CH} _4$
=> $\text{bond dissociation energy of C-H bond in CH} _4 = 99.45Kcal$
Hence, answer is option A.

Given that $\Delta {H} _{f}(H)=218kJ/mol$, express the $H-H$ bond energy in $kcal/mol$:

chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

  1. $52.15$

  2. $911$

  3. $109$

  4. $52153$

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Correct Option: C
Explanation:

$H-H\longrightarrow H+H\quad \quad ;\quad \Delta H$

$\Delta H$ is the bond energy
Therefore, $2\Delta H={ \Delta  } _{ f }H$
            $\therefore \quad \Delta H=\dfrac { { \Delta  } _{ f }\left( H \right)  }{ 2 } $
                             $=\dfrac { 218 }{ 2 } KJ/mol$
                    $\Delta H=109KJ/mol$

The $S-S$ bond energy is: 

$\Delta { H } _{ f }^{ o }({E}{t}-S-{E}{t})=-147kJ/mol$
$\Delta { H } _{ f }^{ o }({E}{t}-S-S-{E}{t})=-202kJ/mol$
$\Delta { H } _{ f }^{ o } S(g)=+223kJ/mol$

chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

  1. $168\ kJ$

  2. $126\ kJ$

  3. $278\ kJ$

  4. $575\ kJ$

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Correct Option: C
Explanation:

${C} _{2}{H} _{5}-S-{C} _{2}{H} _{5}+S(s)\rightarrow {C} _{2}{H} _{5}-S-S-{C} _{2}{H} _{5}$
$\Delta { H } _{ reaction }=\sum { { \Delta H } _{ f(products) }^{ o } } +\sum { \Delta { H } _{ f(reactants) }^{ o } } $
$=(-202)-(-147)=-55kJ$
$\Delta { H } _{ reaction }=\sum { { BE } _{ reactants } } -\sum { { BE } _{ products } } $
$-55=$ Heat of sublimation or enthalpy of atomisation of sulphur$-BE(S-S)$
$-55=223-BE(S-S)$
$BE(S-S)=223+55=278kJ$