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

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 $\text{A}$
Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

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

Bond energy is defined as the energy required to break one mole of bonds in the gaseous state. This is equivalent to the energy released when one mole of bonds is formed from gaseous atoms.

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

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. $309.1 kJ/mol$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $S-F = 1855/6 = 309.1 kJ/mol$
Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

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. $+216 kJ$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 ${ H } _{ 2 }=\dfrac { 433 }{ 2 } =+216.5kJ$

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

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

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

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

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. $76\ kcal$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

C2H4 formation: 2C(s) + 2H2(g) → C2H4; ΔfH = 12.5 kcal. Atomization: 2C(s) → 2C(g); ΔH = 2(171) = 342 kcal. 2H2(g) → 4H(g); ΔH = 2(104.3) = 208.6 kcal. Total atomization = 550.6 kcal. This forms 4 C-H bonds (4×99.3 = 397.2) and 1 C=C bond. BE(C=C) = 550.6 - 397.2 - 12.5 = 140.9 kcal.

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 : $431.37\ kJ\ mol^{-1}$

$C = C$ 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. $-120.0\ kJ\ mol^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Delta H = Sum(BE reactants) - Sum(BE products). Reactants: 1(C=C) + 4(C-H) + 1(H-H) = 606.1 + 4(410.5) + 431.37 = 606.1 + 1642 + 431.37 = 2679.47. Products: 1(C-C) + 6(C-H) = 336.49 + 6(410.5) = 336.49 + 2463 = 2799.49. Delta H = 2679.47 - 2799.49 = -120.02 kJ/mol.

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

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