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Bond enthalpies - class-XII

Description: bond enthalpies
Number of Questions: 58
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Tags: how far? how fast? chemistry energetics and thermochemistry chemical thermodynamics enthalpy changes thermodynamics
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The enthalpies of combustion of carbon and carbon monoxide are -393.5 KJ and -283 KJ respectively the enthalpy of formation of carbon monoxide is :

  1. -676.5 KJ

  2. -110.5 KJ

  3. 110.5 KJ

  4. 676.5 KJ


Correct Option: C

The incorrect statement is :

  1. $ \Delta _{ So| }H^{ 0 }={ \Delta } _{ Lattice }H^{ 0+ }\Delta _{ Hyd }H^{ 0 } $

  2. The enthalpy on dilution is independent on original concentration.

  3. $ N \equiv N > C \equiv N > C \equiv C $ [value of mean bond enthalpy in KJ/mol]

  4. $ H _2O(S) \overset { 1\quad bar }{ \rightleftharpoons } H _2O(I) 273 K $


    $ \Delta U= +Ve; \Delta H = +ve; W= +ve; q = +ve $


Correct Option: B

The enthalpy of tetramerization of $X$ in gas phase $(4X(g)\rightarrow { X } _{ 4 }(g))$ is $-100\ kJ/mol$ at $300\ K$. The enthalpy of vaporisation for liquid $X$ and ${X} _{4}$ are respectively $30\ kJ/mol$ and $72\ kJ/mol$ respectively.
$\Delta S$ for tetramerization of $X$ in liquid phase is $-125\ J/K mol$ at $300\ K$.
What is the $\Delta G$ at $300\ K$ for tetramerization of $X$ in liquid phase?

  1. $-52\ kJ/mol$

  2. $-98\ kJ/mol$

  3. $-14.5\ kJ/mol$

  4. $None\ of\ these$


Correct Option: A

Heat of formation of $2$ moles of ${NH} _{3}(g)$ is $-90kJ$; bond energies of $H-H$ and$N-N$ bonds are $435kJ$ and $390kJ$ ${mol}^{-1}$ respectively. The value of the bond energy of $N\equiv N$ will be:

  1. $-472.5\ kJ$

  2. $-945\ kJ$

  3. $472.5\ kJ$

  4. $945\ kJ$ ${mol}^{-1}$


Correct Option: D
Explanation:

$\Delta { H } _{ reaction }=\sum { { \left( BE \right)  } _{ reactants } } -\sum { { \left( BE \right)  } _{ products } } $
$-90=x+3\times 435-6\times 390$
$x=945kJ$ ${mol}^{-1}$

The standard enthalpy of formation of ${NH} _{3}$ is $-46kJ{ mol }^{ -1 }$. If the enthalpy of formation of ${H} _{2}$ from its atoms is $-436kJ{ mol }^{ -1 }$ and that of ${N} _{2}$ is $-712kJ{ mol }^{ -1 }$, the average bond enthalpy of $N-H$ bond in ${NH} _{3}$ is:

  1. $+1056kJ{ mol }^{ -1 }\quad $

  2. $-1102kJ{ mol }^{ -1 }\quad $

  3. $-964kJ{ mol }^{ -1 }\quad $

  4. $+352kJ{ mol }^{ -1 }\quad $


Correct Option: D
Explanation:

${ H } _{ reaction }=\sum { { BE } _{ reactants } } -\sum { { BE } _{ products } } $
$-46=(\cfrac(712)+\cfrac{3}{2}(436))-3x$
$x=+352\ kJ{ mol }^{ -1 } $

Calculate $ \Delta { H }^{ o }$ of the reaction:

$ { CH } _{ 2 }={ CH } _{ 2 }+3O=O\longrightarrow 2O=C=O+2H-O-H$

The average bond enthelpies of various bond are:

$Bond \quad \quad \quad\quad \quad C-H\quad \quad O=O\quad \quad C=O\quad \quad O-H\quad \quad C=C$
$Bond\ enthalpy$:     414               499               724           460               619
[kJ/mol]

  1. -364kJ

  2. -564kJ

  3. -964kJ

  4. -1654kJ


Correct Option: C

$H _{2}+Cl _{2}\rightarrow 2HCl+44\ K.Cal$. Heat of decomposition of $HCl$ is:

  1. $-44\ K.Cal$

  2. $+44\ K.Cal$

  3. $-22\ K.Cal$

  4. $+22\ K.Cal$


Correct Option: D
Explanation:

$H _2+Cl _2 \longrightarrow 2HCl+44Kcal$

Heat of decomposition of $HCl$ :-
$2HCl \longrightarrow \Delta H _{decomp}=+44 Kcal$
$1 HCl \longrightarrow \Delta H _{decomp}=+22 Kcal$
So, Heat of decomposition of $HCl=+22Kcal$

Two moles of an ideal gas expended isothermally and reversibly from 1 litre to 10 litre at 300 K. The enthalpy change (in kJ) for the process is:

  1. 11.4

  2. -11.4

  3. 0

  4. 4.8


Correct Option: B
Explanation:
Work done in a reversible isothermal process is:

$W = -2.303 \; nRT \; \log{\cfrac{{V} _{f}}{{V} _{i}}} ..... \left( 1 \right)$

Given:-
$n = 2 \text{ moles}$
$T = 300 \; K$
${V} _{f} = 10 \; L$
${V} _{i} = 1 \; L$
$R =$ Gas constant $= 8.314 \; {J}/{K-mol}$

Substituting these values in ${eq}^{n} \left( 1 \right)$, we have

$W = - 2.303 \times 2 \times 8.314 \times 300 \times \log{\cfrac{10}{1}}$

$\Rightarrow W = -11488.285 \; J = -11.4 \; kJ$

Now as we know that,

$\Delta{H} = \Delta{U} + W$

For an isothermal process,

$\Delta{U} = 0$

$\therefore \Delta{H} = W = -11.4 \; kJ$

Hence the enthalpy change for the given process is $-11.4 \; kJ$.

Hence, the correct option is $\text{B}$

The standard enthalpies of n-pentane, isopentane and neopentane are $-35.0,\ -37.0$ and $-40.0$ $K \ cal/mole$ respectively. The most stable isomer of pentane in terms of energy is ____________.

  1. n-pentane

  2. isopentane

  3. neopentane

  4. n-pentane and isopentane


Correct Option: C
Explanation:

The standard enthalpies of n-pentane, isopentane and neopentane are -35.0, -37.0 and -40.0 K.cal/mole respectively. The most stable isomer of pentane in terms of energy is neopentane as it has most negative value of the standard enthalpy.

Calculate P - CI bond enthalpy 
Given : $\Delta f H(PCl _3, g) = 306 KJ/mol;$     $\Delta H _{atomization} (P, s) = 314 KJ / mol;$
$\Delta f H (Cl, g) = 121 KJ / mol$

  1. 123.66 KJ/mol

  2. 371 KJ / mol

  3. 19 KJ/ mol

  4. None of these


Correct Option: A

Given that bond energies of H-H and Cl-Cl are $430$ and $240$ kJ $mol^{-1}$ respectively and $\Delta _fH$ for HCl is $-90$kJ $mol^{-1}$. Bond enthalpy of HCl is?

  1. $290$ kJ $mol^{-1}$

  2. $380$ kJ $mol^{-1}$

  3. $425$ kJ $mol^{-1}$

  4. $245$ kJ $mol^{-1}$


Correct Option: C

Choose the correct order of lattice enthalpy of $LiCl,\ LiF,\ NaCl$ and $NaF$ :

  1. $LiF > NaCl > NaF > LiCl$

  2. $LiF > LiCl > NaF > NaCl$

  3. $LiF > NaF > Nacl > LiCl$

  4. $LiCl > LiF > NaF > NaCl$


Correct Option: B

$100 ml$ of $0.2\ M\ H _{2}SO _{4}$ is reacted with $100\ ml$ of $0.5\ M\ NaOH$ solution. what is the normality of the solution 

  1. 0.3N

  2. 0.8N

  3. 0.1N

  4. 1N


Correct Option: C
Explanation:
$1M-H _2SO _4=2N-H _2SO _4$

100ml of 0.2M 0.2M $H _2SO _4\equiv 100 \times 0.2$ml of 1M $H _2SO _4$

$\equiv 20ml$ of 2N $H _2SO _4$

$\equiv 40ml$ of 2N $H _2SO _4$

$1M NaOH=1N NaOH$

100ml of 0.2M 0.2M $NaOH\equiv 100 \times 0.2$ml of 1M $NaOH$

$\equiv 20ml$ of 1N $NaOH$

neutralisation occurs when acid and base are mixed due to the formation of salt and water.

20ml of 1N $NaOH\equiv $ 20ml of 1N $H _2SO _4$

$20ml \times 1N=200ml \times $ final strength of acid

therefore the normality of solution is $0.1N$

Calculate the average Bond energy of O-F bond in the following reaction :-
$OF _{2(g)}\rightarrow O _{(g)}+2F _{(g)}$
Given :
$OF _{2(g)}\rightarrow OF _{(g)}+F _{(g)}$; $\Delta H$=201 kJ
$OF _{(g)}\rightarrow O _{(g)}+F _{(g)}$; $\Delta H$=199 kJ

  1. 201 kJ

  2. 199 kJ

  3. 200 kJ

  4. 200.9 kJ


Correct Option: B

Which one of the following statement(s) is/are true?

  1. $\Delta E=0$ for combustion of ${ C } _{ 2 }{ H } _{ 6 }(g)$ in a sealed rigid adiabatic container

  2. ${ \Delta } _{ f }{ H }^{ o }(S,monolithic)\ne 0$

  3. If dissociation energy of $C{ H } _{ 4 }(g)$ is $1656kJ/mol$ and ${ C } _{ 2 }{ H } _{ 6 }(g)$ is $2812kJ/mol$, then value of $C-C$ bond energy will be $328kJ/mol$

  4. If ${ \Delta H } _{ f }({ H } _{ 2 }O,g)=-242kJ/mol; { \Delta H } _{ vap }({ H } _{ 2 }O,l)=44kJ/mol$ then ${ \Delta } _{ f }{ H }^{ o }({{OH}^{-}},aq)$ will be $-142kJ/mol$


Correct Option: A,B,C
Explanation:

${ \Delta } _{ f }{ H }^{ o }(S,monolithic)\ne 0$ as it is not elemental form of S.
$\Delta E=0$ for combustion of ${ C } _{ 2 }{ H } _{ 6 }(g)$ in a sealed rigid adiabatic container as in adiabatic process energy exchange is zero.
For 
${ C } _{ 2 }{ H } _{ 6 }(g)$,
$BE _{C-C} = 2812 - 6\times BE _{C-H} = 2812-6\times 1656/4 = 328$kJ/mol

The bond energy in $kcal : mol^{-1}$ of a $C-! ! ! -C$ single bond is approximately :

  1. 1

  2. 10

  3. 100

  4. 1000


Correct Option: C
Explanation:

The bond energy   of a $C-! ! ! -C$ single bond is approximately 100 $kcal : mol^{-1}$. 


It varies from 82 to $157 :kcal : mol^{-1}$.

Hence, option C is correct.

Bond energies can be obtained by using the following relation:
$\Delta H$(reaction) $=\sum$ Bond energy of bonds, broken in the reactants  $- \sum$ Bond energy of bonds, formed in the products

Bond energy depends on three factors:
a. greater is the bond length, lesser is the bond energy
b. bond energy increases with the bond multiplicity
c. bond energy increases with the electronegativity difference between the bonding atoms.

Arrange $N-H$, $O-H$ and $F-H$ bonds in the decreasing order of bond energy:

  1. $F-H > O-H > N-H$

  2. $N-H > O-H > F-H$

  3. $O-H > N-H > F-H$

  4. $F-H > N-H > O-H$


Correct Option: A
Explanation:

Fluorine is more electron-negative than oxygen and oxygen is more electro-negative than nitrogen.

Hence, bond energy between $F-H$ is greater than $O-H$ which is greater than $N-H$.

Bond energies can be obtained by using the following relation:


$\Delta H(reaction) =\sum$ Bond energy of bonds, broken in the reactants $- \sum$ Bond energy of bonds, formed in the products.

Bond energy depends on three factors:
a. greater is the bond length, lesser is the bond energy
b. bond energy increases with the bond multiplicity
c. bond energy increases with the electronegativity difference between the bonding atoms. Which among the following sequences is correct about the bond energy of $C-C$, $C=C$, and $C\equiv C$ bonds?

  1. $C=C > C \equiv C > C-C$

  2. $C\equiv C < C=C < C-C$

  3. $C\equiv C > C=C > C-C$

  4. $C\equiv C > C-C > C=C$


Correct Option: C
Explanation:

bond energy of $C-C  = 347 KJ/mole$
$C=C = 620 KJ/mole$
$C\equiv C =812 KJ/mole$
so order is: 
$C\equiv C > C=C > C-C$

Bond energies can be obtained by using the following relation:
$\Delta H(reaction) =\sum$ Bond energy of bonds, broken in the reactants $- \sum$ Bond energy of bonds, formed in the products
Bond energy depends on three factors:
a. greater is the bond length, lesser is the bond energy
b. bond energy increases with the bond multiplicity
c. bond energy increases with the electronegativity difference between the bonding atoms.Bond energy of different halogen molecules will lie in the sequences of:

  1. $F _2 > Cl _2 > Br _2 > I _2$

  2. $Cl _2 > Br _2 > F _2 > I _2$

  3. $I _2 > Cl _2 > Br _2 > I _2$

  4. $Br _2 > F _2 > I _2 > Cl _2$


Correct Option: B
Explanation:
Bond energy of $F _2$ is low due to lower size of F atoms and strong repulsion between the lone pairs of two fluorine atoms.
Among other halides as size increases bond energy decreases so order is
$Cl _2 > Br _2 > F _2 > I _2$

Bond energies can be obtained by using the following relation:
$\Delta H(reaction) =\sum$ Bond energy of bonds, broken in the reactants $- \sum$ Bond energy of bonds, formed in the products

Bond energy depends on three factors:
a. greater is the bond length, lesser is the bond energy
b. bond energy increases with the bond multiplicity
c. bond energy increases with the electronegativity difference between the bonding atoms.In $CH _4$ molecule.

which of the following statements is correct about the $C-H$ bond energy?

  1. all $C-H$ bonds of methane have same energy.

  2. average of all $C-H$ bond energies is considered.

  3. fourth $C-H$ bond requires highest energy to break.

  4. None of these


Correct Option: B
Explanation:

In the methane, energy needed to break a mole of methane gas into gaseous carbon and hydrogen atoms is +1662 kJ and involves breaking 4 moles of C-H bonds. The average bond energy is therefore +1662/4 kJ, which is +415.5 kJ per mole of bonds.
bond enthalpies give average values of all similar bonds.

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
${ N } _{ 2 }+{ 3H } _{ 2 }\longrightarrow { 2NH } _{ 3 }$
${ \Delta H } _{ r }={ X } _{ 1 }+{ 3X } _{ 2 }-{ 2X } _{ 3 }$

  1. True

  2. False


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:

  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


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. 


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

  1. 1542

  2. 88

  3. 171

  4. 514


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 $A _2,\, AB\, &amp;\, B _2$ 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 $A _2$.

  1. 400 kJ/mole

  2. 300 kJ/mole

  3. 500 kJ/mole

  4. 900 kJ/mole


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.

  1. $-48.9\ kcal$

  2. $-49.2\ kcal$

  3. $-16.5\ kcal$

  4. $-98.1\ kcal$


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:

  1. $1.04\ Kcal$

  2. $10.4\ Kcal$`

  3. $104\ Kcal$

  4. $1040\ Kcal$


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:

  1. $270kcal$

  2. $70kcal$

  3. $200kcal$

  4. $240kcal$


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:

  1. $+99.45kcal$

  2. $-99.45kcal$

  3. $+397.8kcal$

  4. $+198.9kcal$


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 $kcal/mol$:

  1. $52.15$

  2. $911$

  3. $109$

  4. $52153$


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: 

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

  1. $168\ kJ$

  2. $126\ kJ$

  3. $278\ kJ$

  4. $575\ kJ$


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$-BE(S-S)$
$-55=223-BE(S-S)$
$BE(S-S)=223+55=278kJ$

Using the data provided, calculate the multiple bond energy ($kJ{ mol }^{ -1 }$) of $C\equiv  C$ bond in ${C} _{2}{H} _{2}$. That energy is (take the bond energy of a $C-H$ bond as $350kJ{ mol }^{ -1 }$):
$2C(s)+{ H } _{ 2 }(g)\longrightarrow { C } _{ 2 }{ H } _{ 2 }(g);\Delta { H }^{  }=225kJ{ mol }^{ -1 }$
$2C(s)\longrightarrow  2C(g);\Delta { H }^{  }=1410kJ{ mol }^{ -1 }\quad $
${H} _{2}(g)\longrightarrow 2H(g);\Delta { H }^{  }=330kJ{ mol }^{ -1 }\quad $

  1. $1165$

  2. $837$

  3. $865$

  4. $815$


Correct Option: D
Explanation:

$ \Delta H _R = (bond energy) _R - (bond energy) _P$

$225 =( [1410+330]-[\Delta H _{c\equiv c} + 2*350])$
$\therefore \Delta H _{c\equiv c} = 815KJ/mol$

Given that, bond energies of $H-H$ and $Cl-Cl$ ar $430kJ/mol$ and $240kJ/mol$ respectively. $\Delta {H} _{f}$ for $HCl$ is $-90kJ/mol$. Bond enthalpy of $HCl$ is:

  1. $380kJ{ mol }^{ -1 }$

  2. $425kJ{ mol }^{ -1 }$

  3. $245kJ{ mol }^{ -1 }$

  4. $290kJ{ mol }^{ -1 }$


Correct Option: B
Explanation:

$H-H+Cl-Cl\rightarrow 2H-Cl$

${ \Delta H } _{ f }\left( HCl \right) =$ Bond energy of H-H + Bond energy of $Cl-Cl$ - 2(Bond energy of $H-Cl$)
$\therefore $  -90=430+240-(Bond enthalpy of HCl) $\times $ 2
$\therefore $  Bond enthalpy of HCl $=\dfrac { 430+240+90 }{ 2 } =380KJ/mol$

If values of $\Delta { H } _{ f }^{ o }$ of $ICl(g),\, Cl(g),\, I(g)$ are respectively $17.57,\,121.34,\,106.96$ J mol $^{-1}$. The value of $I-Cl$ (bond energy) in J mol $^{-1}$ is:

  1. $17.57$

  2. $210.73$

  3. $35.15$

  4. $106.96$


Correct Option: B
Explanation:
Gas                Values of ${ \Delta H } _{ f }$
$ICl$                     $17.57$
$Cl$                      $121.34$
$I$                         $106.96$
  $I+Cl\rightarrow ICl$
$(g)$  $(g)$      $(g)$
according to Hess less,
$\Delta H={ \Delta H } _{ f }$(reactants) $-\Delta { H } _{ f }$(products)
         $=(106.96+121.34)-(17.57)$
         $=210.73$ J/mol
$\Rightarrow$ Value of $I-Cl$ (bond energy) in J/mol $=210.73$ J/mol

The bond dissociation energies for single covalent bonds formed between carbon and $A,B,C,D$ and $E$ atoms are:

Bond  Bond energy $kcal{ mol }^{ -1 }$
(i) $C-A$ $240$
(ii) $C-B$ $382$
(iii) $C-D$ $276$
(iv) $C-E$ $486$

This indicates that the smallest atom is:

  1. $A$

  2. $B$

  3. $C$

  4. $E$


Correct Option: D
Explanation:

$C-E$ bond has highest bond energy; it means that the covalent bond $C-E$ will be strongest. Smaller is the size of atom, stronger is the covalent bond.

Given the bond dissociation energies below (in kcal/mole), estimate the $\Delta { H }^{ o }$ for the propagation step

${ \left( { CH } _{ 3 } \right) } _{ 2 }CH+{ Cl } _{ 2 }\longrightarrow { \left( { CH } _{ 3 } \right) } _{ 2 }CHCl+Cl$


${ CH } _{ 3 }{ CH } _{ 2 }{ CH } _{ 2 }-H$ $98$
${ \left( { CH } _{ 3 } \right)  } _{ 2 }CH-H$ $95$
$Cl-Cl$ $58$
$H-Cl$ $103$
${ CH } _{ 3 }{ CH } _{ 2 }{ CH } _{ 2 }-Cl$ $81$
${ \left( { CH } _{ 3 } \right)  } _{ 2 }CH-Cl$ $80$
  1. $-30kcal/mol$

  2. $+22kcal/mol$

  3. $-40kcal/mole$

  4. $+45kcal/mol$


Correct Option: A

Given the bond dissociation energies below (in $kcal$ /mole), estimate the $\triangle H^{\circ}$ for the propagation step
$(CH _{3}) _{2} CH + Cl _{2}\rightarrow (CH _{3}) _{2} CHCl + Cl$
$CH _{3} CH _{2}CH _{2} - H \ 98$
$(CH _{3}) _{2} CH - H \ 95$
$Cl - Cl\ 58$
$H-Cl\ 103$
$CH _{3}CH _{2}CH _{2} - Cl\ 81$
$(CH _{3}) _{2} CH - Cl \ 80$

  1. $-30\ kcal/mole$

  2. $+22\ kcal/mole$

  3. $-40\ kcal/mole$

  4. $+45\ kcal/mole$

  5. $-45\ kcal/mole$


Correct Option: A

If the bond energies of $H-H,\ Br-Br$ and $H-Br$ are 433, 192 and 364 $kJ \, mol^{-1}$ respectively, $\Delta H$ for the reaction $H _{2(g)}+BR _{2(g)}\rightarrow 2HBr _{(g)}$ is:

  1. -261 kJ

  2. +103 kJ

  3. +261 kJ

  4. -103 kJ


Correct Option: D

$NO(g) + O _{3}(g)\rightarrow NO _{2}(g) + O _{2}(g)\ triangle H = -198.9\ kJ/mol$
$O _{3}(g) \rightarrow 3/2\ O _{2}(g) \ \triangle H = -142.3\ kJ/mol$
$O _{2}(g) \rightarrow 2O(g) \ \triangle H = +495.0\ kJ/mol$

The enthalpy change $(\triangle H)$ for the following reaction is
$NO(g) + O(g)\rightarrow NO _{2}(g)$

  1. $-304.1\ kJ/ mol$

  2. $+304.1\ kJ/ mol$

  3. $-403.1\ kJ/ mol$

  4. $+403.1\ kJ/ mol$


Correct Option: A
Explanation:

$NO(g)+O _{ 3 }(g)\rightarrow NO _{ 2 }(g)+O _{ 2 }(g)\quad ;\quad \Delta H=-198.9kJ/mol-----(i)\ \quad \quad \quad \quad \quad \quad \quad O _{ 3 }(g)\rightarrow \cfrac { 3 }{ 2 } O _{ 2 }(g)\quad ;\quad \Delta H=-142.3kJ/mol\ \cfrac { 3 }{ 2 } O _{ 2 }(g)\rightarrow O _{ 3 }(g)\quad ;\quad \Delta H=142.3kJ/mol-----(ii)\ \quad \quad \quad \quad \quad \quad \quad \quad O _{ 2 }(g)\rightarrow 2O(g)\quad ;\quad \Delta H=+495.0kJ/mol\ \quad \quad \quad \quad \quad \quad \quad \quad 2O(g)\rightarrow O _{ 2 }(g)\quad ;\quad \Delta H=-495.0kJ/mol\ O(g)\rightarrow \cfrac { 1 }{ 2 } O _{ 2 }(g)\quad ;\quad \Delta H=-\cfrac { 495.0 }{ 2 } kJ/mol-----(iii)\ Adding\quad (i),\quad (ii)\quad and\quad (iii),\ NO(g)+O(g)\rightarrow NO _{ 2 }(g)\quad ;\quad \Delta H=(-198.9+142.3-\cfrac { 495.0 }{ 2 } )kJ/mol\ \therefore \Delta H=-304.1kJ/mol$

$\overset { \underset { | }{ H }  }{ \underset { \overset { | }{ H }  }{ C }  } =\overset { \underset { | }{ H }  }{ \underset { \overset { | }{ H }  }{ C }  } +H-H\rightarrow H-\overset { \underset { | }{ H }  }{ \underset { \overset { | }{ H }  }{ C }  } -\overset { \underset { | }{ H }  }{ \underset { \overset { | }{ H }  }{ C }  } -H$
From the following bond energies:
$H-H$ bond energy: $431.37kJ\quad { mol }^{ -1 }$
$C=C$ bond energy: $606.10kJ\quad { mol }^{ -1 }\quad $
$C-C$ bond energy: $336.49kJ\quad { mol }^{ -1 }$
$C-H$ bond energy: $410.50kJ\quad { mol }^{ -1 }$
Enthalpy for the reaction will be:

  1. $553.0kJ\quad { mol }^{ -1 }$

  2. $1523.6kJ\quad { mol }^{ -1 }$

  3. $-243.6kJ\quad { mol }^{ -1 }$

  4. $-120.0kJ\quad { mol }^{ -1 }$


Correct Option: D
Explanation:
$\overset { \underset { | }{ H }  }{ \underset { \overset { | }{ H }  }{ C }  } =\overset { \underset { | }{ H }  }{ \underset { \overset { | }{ H }  }{ C }  } +H-H\rightarrow H-\overset { \underset { | }{ H }  }{ \underset { \overset { | }{ H }  }{ C }  } -\overset { \underset { | }{ H }  }{ \underset { \overset { | }{ H }  }{ C }  } -H$
Enthalpy of the reaction $=-$[$6\times(C-H)$ bond energy $+1(C-C)$ bond energy $-(H-H$ bond energy$)-(C=C)$ bond energy $-4(C-H)$ bond energy]
$=-\left[ 6\times 410.50+1\times 336.49-431.37-606.10-4\times 410.50 \right] $
$=-120.0kJ{ mol }^{ -1 }$

If, $C(s)+2H _2(g)\rightarrow CH _4(g);       \triangle H= -X _1 kcal$ 
   $C(g)+4H(g)\rightarrow CH _4(g);            \triangle H = -X _2 kcal$
   $CH _4(g) \rightarrow CH _3(g)+H(g); \triangle H = +Y kcal$
The average bond energy of C-Hbond in kcal $mol^{-1}$ is :

  1. $\frac{X _1}{4}$

  2. Y

  3. $\frac{X _2}{4}$

  4. $X _1$


Correct Option: C

Which is the correct order of bond energy of single, double and triple bonds between carbon atoms?

  1. $C-C > C= C > C \equiv C$

  2. $C= C > C \equiv C > C-C$

  3. $C \equiv C > C-C > C =C$

  4. $C \equiv C > C = C > C-C$


Correct Option: D
Explanation:

Carbon has all 4 valence electrons. So it easily bonds with other carbon atoms to form long chains Carbon atoms. Apart from single bond carbon also forms double bonds in compounds like alkenes or triple bonds in compounds like alkynes.

A carbon-carbon bond is very strong. The carbon to carbon triple bond has dissociation energy of $348 \frac{kJ}{mol}$, followed by a double bond with $614 \frac{kJ}{mol}$ and single bond having $839 \frac{kJ}{mol}$.

 The correct order of bond energy strength is option D.

Bond energies of H - H and CI - CI are $430 \ kJ mol^{-1}$ and $242 \ kJ mol^{-1}$ respectively. $\Delta H _f$ for HCl is $91 \ kJ mol^{-1}$ . What will be the bond energy of H - Cl bond (per mole value)?

  1. 672 kJ

  2. 182 kJ

  3. 245 kJ

  4. 88 kJ


Correct Option: C
Explanation:

$\frac{1}{2} H _2 + \frac{1}{2} Cl _2 \rightarrow HCl$
$\Delta H _f(HCl) = 91 \ kJ \ mol^{-1}$, H - H = 430 kJ $mol^{-1}$, Cl - Cl = 242 kJ $mol^{-1}$
$91 = \frac{1}{2} \times 430 + \frac{1}{2} \times 242 - B.E. (H - Cl)$
B.E. (H-Cl) = 245 kJ

Which of the following relationships is not correct?

  1. $\Delta H = \Delta E + \Delta n _gRT$

  2. $\Delta H _{sub} = \Delta H _{fusion} + \Delta H _{vap}$

  3. $\Delta H _r^0 = \sum H _{f(reactants)}^0 - \sum H _{f(products)}^0$

  4. $\Delta H _r^0 = \sum B.E. $ of reactants $- \sum B.E. $ of products


Correct Option: C
Explanation:

From definition of Heat of Formation and Hess' law, $\Delta H _r^0 = \sum H _{f(products)}^0 - \sum H _{f(reactants)}^0$

Bond energies of some bonds are given below:
Cl-Cl = 242.8 kJ $mol^{-1}$, H-Cl = 431.8 kJ $mol^{-1}$,
O-H = 464  kJ $mol^{-1}$, O=O = 442 kJ $mol^{-1}$
Using the B.E.s given, calculate $\Delta H$ for the given reaction: $2Cl _2 + 2H _2O \rightarrow 4HCl + O _2$ 

  1. 906 kJ $mol^{-1}$

  2. 172.4 kJ $mol^{-1}$

  3. 198.8 kJ $mol^{-1}$

  4. 442 kJ $mol^{-1}$


Correct Option: B
Explanation:

$2CI _2 + 2H _2O \rightarrow 4HCI + O _2$   

From Hess' Law,  $\Delta H$ = B.E. of (2 X CI - CI) + (2 X 2 X O - H) - (4 X H - CI) + (O = O) 
                                      = 2 X 242.8 + 4 X 464 - 4 X 431.8 - 442               

                                      = 172.4 kJ $mol^{-1}$

The enthalpy of dissociation of $PH _3$ is $954$ kJ/mol and that of $P _2H _4$ is $1.485$ MJ/mol. What is the bond enthalpy of $P-P$ bond? 

  1. 213 kJ/mol

  2. 413 kJ/mol

  3. 200 kJ/mol

  4. Given data is incorrect


Correct Option: A
Explanation:
$PH _3\longrightarrow P+3H\quad \Delta H =954\ KJ\ mol^{-1}$

$\therefore \ 3\Delta H _{P-H}=\Delta ^rH\ \Rightarrow \ \Delta H _{P-H}=\dfrac {954}{3}KJ\ mol^{-1}$

Also given,
Enthalpy of dissociation of $P _2H _4$ is $1.485\ MJ/mol=1485\ KJ\ mol^{-1}$

$\therefore \ P _2H _4\longrightarrow 2P+4H\ \Delta ^rH=1485\ KJ\ mol^{-1}$

$\Rightarrow \ 4\Delta H _{P-H}+\Delta H _{P-P}=\Delta ^rH$

$\Rightarrow \ \Delta H _{P-P}= \Delta^r H-4\Delta H _{P-H}=1485-4\times \dfrac {954}{3}=213\ KJ\ mol^{-1}$

$\Rightarrow \ \Delta H _{P-P}=213\ KJ\ mol^{-1}$

Thus bond enthalpy of $P-P$ bond $=213\ KJ\ mol^{-1}$

Hence, the correct option is $\text{A}$

Amount of energy required to break a specific covalent bond is called:

  1. Bond dissociation energy


  2. Bond energy

  3. Bond enthalpy

  4. All of the above


Correct Option: D
Explanation:

The energy required to break a specific covalent bond in one mole of gaseous molecules is called the bond energy or the bond dissociation energy or bond enthalpy.


Hence, the correct option is $D$ 

The bond energy of $H _2$ is $104.3 : kcal : mol^{-1}$. It means that:

  1. 104.3 kcal heat is needed to break up $'! N'$ bonds in $N$ moleules of $H _2$

  2. 104.3 kcal heat is needed to break up $6.023\times 10^{23}$ molecules into $1.2046\times 10^{24}$ of H

  3. 104.3 kcal heat is evolved during combination of $2N$ atoms of H to form $N$ molecules of $H _2$

  4. All of the above


Correct Option: D

The average, $S - F$ bond energy in $SF _6$ if the $\Delta H^{\circ} _f$ value are $-1100, +275$ and $+80 kJ/mol$ respectively for $SF _6(g),$ S(g) and F(g) is

  1. $390.1 kJ/mol$

  2. $103.9 kJ/mol$

  3. $903.1 kJ/mol$

  4. $309.1 kJ/mol$


Correct Option: D
Explanation:
As we know,
Heat of a reaction = Bond energy of reactants - Bond energy of products
here,
$S(s) +3F _2(g) \rightarrow SF _6(g)$
so heat of reaction = 
$-1100 = 275 + 6*80$ - bond energy of $SF _6$ 
bond energy of $SF _6$ $= 1855$
so bond energy of $S-F = 1855/6 = 309.1 kJ/mol$

Bond energy of hydrogen gas is $-433 kJ$. How much is the bond dissociation energy of $0.5 mole$ of hydrogen gas?

  1. $-433 kJ$

  2. $+433 kJ$

  3. $-216 kJ$

  4. $+216 kJ$


Correct Option: D
Explanation:

Bond dissociation energy $= -$ bond formation energy
Given, bond energy of hydrogen $= -433 kJ$
$\therefore$ Bond dissociation energy of one mole ${ H } _{ 2 } = 433 kJ$
$\therefore$ Bond dissociation energy of $0.5$ mole ${ H } _{ 2 }=\dfrac { 433 }{ 2 } =+216.5kJ$

A vessel contains 100 litres of a liquid x. Heat is supplied to the liquid in such a fashion that, heat gives a change in enthalpy. The volume of the liquid increases by 2 litres. If the external pressure is one atm, and 202.6 Joules of heat supplied, then [U $\rightarrow$ total internal energy] :

  1. $\Delta U=0, \Delta H=0$

  2. $\Delta U=+202.6 J, \Delta H=+202.6 J$

  3. $\Delta U=-202.6 J, \Delta H=-202.6 J$

  4. $\Delta U=0, \Delta H=+202.6 L $


Correct Option: D

Enthalpy of polymerisation of ethylene, as represented by the reaction, $ nCH _2 = CH _2 \rightarrow {(-CH _2- CH _2-)} _n   $ is -100kJ per mole of ethylene.Given bond enthalpy of $ C = C $ bond is 600 kJ$ mol^{-1} $ , enthalpy of $ C - C $ bond (in kJ mol) will be :

  1. 2940 kcal $ mol^{-1} $

  2. 350 kJ $ mol^{-1} $

  3. 700 kJ $ mol^{-1} $

  4. 1470 kcal $ mol^{-1} $


Correct Option: B,D
Explanation:

$ nCH _2 = CH _2 \rightarrow (-CH _2 - CH _2- ) _n    \Delta H $= 100 KJ/mole
Double bond of ethylene converted to two single bonds.
$ \Delta H $ = - 100 =$ B.E. _{C=C} - 2 B.E. _{C-C}$ 
$ \implies $ - 100 = 600 - $ 2 \times B.E. _{C-C} \implies B.E. _{C-C}$ = 350 KJ/mole = 1470kcal$mol^{-1}$

The first and second dissociation constant of an acid ${ H } _{ 2 }A$ are $1.0\ \times \ { 10 }^{ -5 }$ and $5.0\ \times \ { 10 }^{ -10 }$ respectively. The over all dissociation constant of the acid will be:

  1. $5.0\ \times \ { 10 }^{ -5 }$

  2. $5.0\ \times \ { 10 }^{ 15 }$

  3. $5.0\ \times \ { 10 }^{ -15 }$

  4. $0.2\ \times \ { 10 }^{ 5 }$


Correct Option: C
Explanation:

$H _2A\overset {K _1}{\rightleftharpoons} HA^-+H^+$


$\Rightarrow K _1=\cfrac {[HA^-][H^+]}{[H _2A]}$  $\longrightarrow (1)$


$HA^-\overset {K _2}{\rightleftharpoons} H^++A^{2-}$

$\Rightarrow K _2=\cfrac {[H^+][A^{2-}]}{[HA^-]}$    $\longrightarrow (2)$

Overall dissociation constant $K$

$\Rightarrow K=\cfrac {[H^+]^2[A^{2-}]}{[H _2A]}=K _1\times K _2$

$=1\times 10^{-5}\times 5\times 10^{-10}$

$=5\times 10^{-15}$ .

$\triangle H _{f} (C _{2}H _{4}) = 12.5\ kcal$

Heat of atomisation of $C = 171\ kcal$
Bond energy of $H _{2} = 104.3\ kcal$
Bond energy $C - H = 99.3\ kcal$

What is $C = C$ bond energy?

  1. $140.9\ kcal$

  2. $49\ kcal$

  3. $40\ kcal$

  4. $76\ kcal$


Correct Option: A
$\overset { \underset { | }{ H }  }{ \underset { \overset { | }{ H }  }{ C }  }=\overset { \underset { | }{ H }  }{ \underset { \overset { | }{ H }  }{ C }  } +H-H\rightarrow H-\overset { \underset { | }{ H }  }{ \underset { \overset { | }{ H }  }{ C }  } -\overset { \underset { | }{ H }  }{ \underset { \overset { | }{ H }  }{ C }  } -H $

From the following bond energies:

$H - H$ bond energy : $431.37\ kJ\ mol^{-1}$

$C = C$ bond energy : $606.10\ kJ\ mol^{-1}$

$C - C$ bond energy : $336.49\ kJ\ mol^{-1}$

$C - H$ bond energy : $410.50\ kJ\ mol^{-1}$

Enthalpy for the reactions, will be?
  1. $+553.0\ kJ\ mol^{-1}$

  2. $+1523.6\ kJ\ mol^{-1}$

  3. $-243.6\ kJ\ mol^{-1}$

  4. $-120.0\ kJ\ mol^{-1}$


Correct Option: D

Dissociation of water takes place in two steps:
$H _2O \rightarrow H^+ + OH^-$; $\Delta H$ = +497.8 kJ
$OH^- \rightarrow H^+ + O^{2-}$; $\Delta H$ = +428.5 kJ
What is the bond energy of O - H bond?

  1. 463.15 kJ $mol^{-1}$

  2. 428.5 kJ $mol^{-1}$

  3. 69.3 kJ $mol^{-1}$

  4. 926.3 kJ $mol^{-1}$


Correct Option: A
Explanation:

In given 2 reactions, $\Delta H$ is basically representing bond energies of H-O bond. So, our answer should be average of these two $\Delta H$ values.
Hence, Average of two bond dissociation energies:    $\frac{497.8 + 428.5}{2}$ = 463.15kJ $mol^{-1}$

The dissociation energy of $CH _4$ and $C _2H _6$ are respectively 360 and 620 kcal /mole. the bond energy $C-C$ is:

  1. 260 kcal /mole

  2. 180 kcal /mole

  3. 130 kcal / mole

  4. 80 kcal / mole


Correct Option: D
Explanation:
Dissociation of energy of $CH _4$ base
=$\cfrac {\text{dissociation energy of }{CH _4}}{4}=\cfrac {360}{4}=90Kcal/mole$

Dissociation energy of $C _2H _6$

620=1(B.E of C-C bond+ 6 B.E of C-H bond)

620=B.E of C-C bond +6 $\times $90

B.E of C-C bond=620-540=80 kcal / mole

The heat of neutralisation of $HCl$ by $NaOH$ is -55.9 KJ/mole. If the heat of neutralisation of $HCN$ by$ NaOH$ is -12.1 KJ/mole, the energy of dissociation of $HCN$ is:

  1. -43.8 KJ

  2. 43.8 KJ

  3. 68 KJ

  4. -68 KJ


Correct Option: B
Explanation:

Heat of dissociation $=(55.9-12.1)=43.8\ KJ$

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