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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

Which is the correct order of bond energy of single, double and triple bonds between carbon atoms?

  1. $C-C > C= C > C \equiv C$
  2. $C= C > C \equiv C > C-C$
  3. $C \equiv C > C-C > C =C$
  4. $C \equiv C > C = C > C-C$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Carbon has all 4 valence electrons. So it easily bonds with other carbon atoms to form long chains Carbon atoms. Apart from single bond carbon also forms double bonds in compounds like alkenes or triple bonds in compounds like alkynes.

A carbon-carbon bond is very strong. The carbon to carbon triple bond has dissociation energy of $348 \frac{kJ}{mol}$, followed by a double bond with $614 \frac{kJ}{mol}$ and single bond having $839 \frac{kJ}{mol}$.

 The correct order of bond energy strength is option D.

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

Which of the following relationships is not correct?

  1. $\Delta H = \Delta E + \Delta n _gRT$
  2. $\Delta H _{sub} = \Delta H _{fusion} + \Delta H _{vap}$
  3. $\Delta H _r^0 = \sum H _{f(reactants)}^0 - \sum H _{f(products)}^0$
  4. $\Delta H _r^0 = \sum B.E. $ of reactants $- \sum B.E. $ of products
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

From definition of Heat of Formation and Hess' law, $\Delta H _r^0 = \sum H _{f(products)}^0 - \sum H _{f(reactants)}^0$

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

Amount of energy required to break a specific covalent bond is called:

  1. Bond dissociation energy

  2. Bond energy

  3. Bond enthalpy

  4. All of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The energy required to break a specific covalent bond in one mole of gaseous molecules is called the bond energy or the bond dissociation energy or bond enthalpy.


Hence, the correct option is $D$ 

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.