Chemistry

Thermochemistry and Equilibrium

121 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

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.

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 $\triangle H^{\circ}$ 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. $-45\ kcal/mole$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is the same question as 454015. Propagation step: (CH3)2CH-H + Cl-Cl → (CH3)2CH-Cl + H-Cl. ΔH = (95 + 58) - (80 + 103) = -30 kcal/mol. The table gives bond dissociation energies for similar compounds to estimate the reaction enthalpy.

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

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

Delta H = Sum(BE reactants) - Sum(BE products). Delta H = [BE(H-H) + BE(Br-Br)] - 2*BE(H-Br) = [433 + 192] - 2(364) = 625 - 728 = -103 kJ.

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

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

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. $+403.1\ kJ/ mol$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

$\overset { \underset { | }{ H }  }{ \underset { \overset { | }{ H }  }{ C }  } =\overset { \underset { | }{ H }  }{ \underset { \overset { | }{ H }  }{ C }  } +H-H\rightarrow H-\overset { \underset { | }{ H }  }{ \underset { \overset { | }{ H }  }{ C }  } -\overset { \underset { | }{ H }  }{ \underset { \overset { | }{ H }  }{ C }  } -H$
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. $-120.0kJ\quad { mol }^{ -1 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
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] $
$=-120.0kJ{ mol }^{ -1 }$
Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

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. $X _1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The average C-H bond energy relates to atomization of methane. C(g) + 4H(g) → CH4(g) releases X2 kJ (4 bonds form). So average C-H bond energy = X2/4 kJ/mol. X1 is the formation from elements, not atomization from gaseous atoms.

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

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 $mol^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $mol^{-1}$

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\text{A}$