Tag: energetics and thermochemistry

Quizzes (6) Questions (104)

Questions Related to energetics and thermochemistry

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

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:

  1. $A$
  2. $B$
  3. $C$
  4. $E$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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.