Chemistry

Thermochemistry and Equilibrium

110 Questions

Thermochemistry and equilibrium problems focus on calculating bond energies, lattice enthalpy, and the Born Haber cycle. These concepts are vital for scoring well in chemistry sections. Regular practice ensures a clear understanding of energy changes in reactions.

Bond energy calculationsBorn Haber cycleLattice enthalpyEnthalpy of solutionThermochemical equations

Thermochemistry and Equilibrium Questions

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.

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

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

As we know,

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

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

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

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

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

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

$AB,\, A _2$ 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 $A _2$.

  1. 400 kJ/mole

  2. 300 kJ/mole

  3. 500 kJ/mole

  4. 900 kJ/mole
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $A _2= 1x = 400kJmol^{-1}$

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

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. $-98.1\ kcal$
Reveal answer Fill a bubble to check yourself
A Correct answer
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$=-98.1-(-45.2)=52.9\ kcal$

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

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. $1040\ Kcal$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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. $240kcal$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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. $+198.9kcal$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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. $52153$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 } $
                             $=\dfrac { 218 }{ 2 } KJ/mol$
                    $\Delta H=109KJ/mol$

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

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. $575\ kJ$
Reveal answer Fill a bubble to check yourself
C Correct answer
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$-BE(S-S)$
$-55=223-BE(S-S)$
$BE(S-S)=223+55=278kJ$

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

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. $815$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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

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. $290kJ{ mol }^{ -1 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $=\dfrac { 430+240+90 }{ 2 } =380KJ/mol$

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

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. $106.96$
Reveal answer Fill a bubble to check yourself
B Correct answer
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
Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

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. $+45kcal/mol$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Delta H = Energy of bonds broken - Energy of bonds formed. Bonds broken: (CH3)2CH-H (95) + Cl-Cl (58) = 153. Bonds formed: (CH3)2CH-Cl (80) + H-Cl (103) = 183. Delta H = 153 - 183 = -30 kcal/mol.