Tag: energetics and thermochemistry

Questions Related to energetics and thermochemistry

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?

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

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

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

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

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

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Correct Option: C

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

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

  1. $LiF > NaCl > NaF > LiCl$

  2. $LiF > LiCl > NaF > NaCl$

  3. $LiF > NaF > Nacl > LiCl$

  4. $LiCl > LiF > NaF > NaCl$

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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 

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

  1. 0.3N

  2. 0.8N

  3. 0.1N

  4. 1N

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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

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

  1. 201 kJ

  2. 199 kJ

  3. 200 kJ

  4. 200.9 kJ

Show answer
Correct Option: B

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

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

  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 $-142kJ/mol$

Show answer
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 :

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

  1. 1

  2. 10

  3. 100

  4. 1000

Show answer
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:

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

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

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

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

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

Show answer
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?

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

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

Show answer
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:

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

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

Show answer
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?

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

  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

Show answer
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.