Tag: chemistry

Questions Related to chemistry

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

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

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

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

  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. $C \equiv C > C = C > C-C$

Show answer
Correct Option: D
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.

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)?

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

  1. 672 kJ

  2. 182 kJ

  3. 245 kJ

  4. 88 kJ

Show answer
Correct Option: C
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

Which of the following relationships is not correct?

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

  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

Show answer
Correct Option: C
Explanation:

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

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: $2Cl _2 + 2H _2O \rightarrow 4HCl + O _2$ 

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

  1. 906 kJ $mol^{-1}$

  2. 172.4 kJ $mol^{-1}$

  3. 198.8 kJ $mol^{-1}$

  4. 442 kJ $mol^{-1}$

Show answer
Correct Option: B
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 $mol^{-1}$

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? 

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

  1. 213 kJ/mol

  2. 413 kJ/mol

  3. 200 kJ/mol

  4. Given data is incorrect

Show answer
Correct Option: A
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 $\text{A}$

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

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

  1. Bond dissociation energy


  2. Bond energy

  3. Bond enthalpy

  4. All of the above

Show answer
Correct Option: D
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 $D$ 

The bond energy of $H _2$ is $104.3 : kcal : mol^{-1}$. It means that:

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

  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

Show answer
Correct Option: D