Tag: bond energies and enthalpy changes

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

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. $LiCl > LiF > NaF > NaCl$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Lattice enthalpy is inversely proportional to inter-ionic distance. LiF has the smallest ions and thus the highest lattice energy, followed by LiCl, NaF, and NaCl (larger ions).

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

$100 ml$ 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
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $0.1N$
Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The overall reaction for atomization of OF2 into one oxygen and two fluorine atoms is the sum of the two given steps, so its total enthalpy change is 201 + 199 = 400 kJ. Since OF2 contains two O-F bonds, the average bond energy is obtained by dividing the total enthalpy by two, yielding 400 / 2 = 200 kJ.

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

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 $-142kJ/mol$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
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

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

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. $F-H > N-H > O-H$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $N-H$.

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

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. $C\equiv C > C-C > C=C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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. $Br _2 > F _2 > I _2 > Cl _2$
Reveal answer Fill a bubble to check yourself
B Correct answer
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
$Cl _2 > Br _2 > F _2 > I _2$
Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.