Tag: energetics and thermochemistry

Questions Related to energetics and thermochemistry

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 $

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

  1. $1165$

  2. $837$

  3. $865$

  4. $815$

Show answer
Correct Option: D
Explanation:

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

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

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:

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

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

  2. $425kJ{ mol }^{ -1 }$

  3. $245kJ{ mol }^{ -1 }$

  4. $290kJ{ mol }^{ -1 }$

Show answer
Correct Option: B
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 $=\dfrac { 430+240+90 }{ 2 } =380KJ/mol$

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:

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

  1. $17.57$

  2. $210.73$

  3. $35.15$

  4. $106.96$

Show answer
Correct Option: B
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

The bond dissociation energies for single covalent bonds formed between carbon and $A,B,C,D$ and $E$ atoms are:

Bond  Bond energy $kcal{ mol }^{ -1 }$
(i) $C-A$ $240$
(ii) $C-B$ $382$
(iii) $C-D$ $276$
(iv) $C-E$ $486$

This indicates that the smallest atom is:

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

  1. $A$

  2. $B$

  3. $C$

  4. $E$

Show answer
Correct Option: D
Explanation:

$C-E$ bond has highest bond energy; it means that the covalent bond $C-E$ will be strongest. Smaller is the size of atom, stronger is the covalent bond.

Given the bond dissociation energies below (in kcal/mole), estimate the $\Delta { H }^{ o }$ for the propagation step

${ \left( { CH } _{ 3 } \right) } _{ 2 }CH+{ Cl } _{ 2 }\longrightarrow { \left( { CH } _{ 3 } \right) } _{ 2 }CHCl+Cl$


${ CH } _{ 3 }{ CH } _{ 2 }{ CH } _{ 2 }-H$ $98$
${ \left( { CH } _{ 3 } \right)  } _{ 2 }CH-H$ $95$
$Cl-Cl$ $58$
$H-Cl$ $103$
${ CH } _{ 3 }{ CH } _{ 2 }{ CH } _{ 2 }-Cl$ $81$
${ \left( { CH } _{ 3 } \right)  } _{ 2 }CH-Cl$ $80$
chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

  1. $-30kcal/mol$

  2. $+22kcal/mol$

  3. $-40kcal/mole$

  4. $+45kcal/mol$

Show answer
Correct Option: A

Given the bond dissociation energies below (in $kcal$ /mole), estimate the $\triangle H^{\circ}$ for the propagation step
$(CH _{3}) _{2} CH + Cl _{2}\rightarrow (CH _{3}) _{2} CHCl + Cl$
$CH _{3} CH _{2}CH _{2} - H \ 98$
$(CH _{3}) _{2} CH - H \ 95$
$Cl - Cl\ 58$
$H-Cl\ 103$
$CH _{3}CH _{2}CH _{2} - Cl\ 81$
$(CH _{3}) _{2} CH - Cl \ 80$

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

  1. $-30\ kcal/mole$

  2. $+22\ kcal/mole$

  3. $-40\ kcal/mole$

  4. $+45\ kcal/mole$

  5. $-45\ kcal/mole$

Show answer
Correct Option: A

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:

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

  1. -261 kJ

  2. +103 kJ

  3. +261 kJ

  4. -103 kJ

Show answer
Correct Option: D

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

The enthalpy change $(\triangle H)$ for the following reaction is
$NO(g) + O(g)\rightarrow NO _{2}(g)$

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

  1. $-304.1\ kJ/ mol$

  2. $+304.1\ kJ/ mol$

  3. $-403.1\ kJ/ mol$

  4. $+403.1\ kJ/ mol$

Show answer
Correct Option: A
Explanation:

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

$\overset { \underset { | }{ H }  }{ \underset { \overset { | }{ H }  }{ C }  } =\overset { \underset { | }{ H }  }{ \underset { \overset { | }{ H }  }{ C }  } +H-H\rightarrow H-\overset { \underset { | }{ H }  }{ \underset { \overset { | }{ H }  }{ C }  } -\overset { \underset { | }{ H }  }{ \underset { \overset { | }{ H }  }{ C }  } -H$
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:

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

  1. $553.0kJ\quad { mol }^{ -1 }$

  2. $1523.6kJ\quad { mol }^{ -1 }$

  3. $-243.6kJ\quad { mol }^{ -1 }$

  4. $-120.0kJ\quad { mol }^{ -1 }$

Show answer
Correct Option: D
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] $
$=-120.0kJ{ mol }^{ -1 }$

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 :

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

  1. $\frac{X _1}{4}$

  2. Y

  3. $\frac{X _2}{4}$

  4. $X _1$

Show answer
Correct Option: C