Tag: chemistry

Questions Related to chemistry

Bond energies can be obtained by using the following relation:


$\Delta H(reaction) =\sum$ Bond energy of bonds, broken in the reactants $- \sum$ Bond energy of bonds, formed in the products.

Bond energy depends on three factors:
a. greater is the bond length, lesser is the bond energy
b. bond energy increases with the bond multiplicity
c. bond energy increases with the electronegativity difference between the bonding atoms. Which among the following sequences is correct about the bond energy of $C-C$, $C=C$, and $C\equiv C$ bonds?

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

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

  2. $C\equiv C < C=C < C-C$

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

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

Show answer
Correct Option: C
Explanation:

bond energy of $C-C  = 347 KJ/mole$
$C=C = 620 KJ/mole$
$C\equiv C =812 KJ/mole$
so order is: 
$C\equiv C > C=C > C-C$

Bond energies can be obtained by using the following relation:
$\Delta H(reaction) =\sum$ Bond energy of bonds, broken in the reactants $- \sum$ Bond energy of bonds, formed in the products
Bond energy depends on three factors:
a. greater is the bond length, lesser is the bond energy
b. bond energy increases with the bond multiplicity
c. bond energy increases with the electronegativity difference between the bonding atoms.Bond energy of different halogen molecules will lie in the sequences of:

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

  1. $F _2 > Cl _2 > Br _2 > I _2$

  2. $Cl _2 > Br _2 > F _2 > I _2$

  3. $I _2 > Cl _2 > Br _2 > I _2$

  4. $Br _2 > F _2 > I _2 > Cl _2$

Show answer
Correct Option: B
Explanation:
Bond energy of $F _2$ is low due to lower size of F atoms and strong repulsion between the lone pairs of two fluorine atoms.
Among other halides as size increases bond energy decreases so order is
$Cl _2 > Br _2 > F _2 > I _2$

Bond energies can be obtained by using the following relation:
$\Delta H(reaction) =\sum$ Bond energy of bonds, broken in the reactants $- \sum$ Bond energy of bonds, formed in the products

Bond energy depends on three factors:
a. greater is the bond length, lesser is the bond energy
b. bond energy increases with the bond multiplicity
c. bond energy increases with the electronegativity difference between the bonding atoms.In $CH _4$ molecule.

which of the following statements is correct about the $C-H$ bond energy?

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

  1. all $C-H$ bonds of methane have same energy.

  2. average of all $C-H$ bond energies is considered.

  3. fourth $C-H$ bond requires highest energy to break.

  4. None of these

Show answer
Correct Option: B
Explanation:

In the methane, energy needed to break a mole of methane gas into gaseous carbon and hydrogen atoms is +1662 kJ and involves breaking 4 moles of C-H bonds. The average bond energy is therefore +1662/4 kJ, which is +415.5 kJ per mole of bonds.
bond enthalpies give average values of all similar bonds.

If BE (bond energy) of N $\equiv $ N bond is ${ X } _{ 1 }$ that of H H bond is  ${ X } _{ 2 }$ and N H bond is ${ X } _{ 1 }$ then enthalpy change of the reaction is
${ N } _{ 2 }+{ 3H } _{ 2 }\longrightarrow { 2NH } _{ 3 }$
${ \Delta H } _{ r }={ X } _{ 1 }+{ 3X } _{ 2 }-{ 2X } _{ 3 }$

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

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

As we know,

Heat of reaction = bond energy of reactant - bond energy of product
${ N } _{ 2 }+{ 3H } _{ 2 }\longrightarrow { 2NH } _{ 3 }$
so
${ \Delta H } _{ r }={ X } _{ 1 }+{ 3X } _{ 2 }-{ 2\times3X } _{ 3 }$
${ \Delta H } _{ r }={ X } _{ 1 }+{ 3X } _{ 2 }-{ 6X } _{ 3 }$

The bond energy is the energy required to:

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

  1. dissociation one liter of substance

  2. dissociate bond of 1 kg of substance

  3. break one mole of covalently bonded gas molecules

  4. break bonds in one gram of substance

Show answer
Correct Option: C
Explanation:

Bond energy is the measure of bond strength in a chemical bond. It is the energy or heat required to break one mole of molecules into their individual atoms.

The enthalpy change for the following reaction is 514 kJ. Calculate the average Cl - F bond energy. 


   $ClF _3(g)\rightarrow Cl(g)+3:F(g)$

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

  1. 1542

  2. 88

  3. 171

  4. 514

Show answer
Correct Option: C
Explanation:
   $ClF _3(g)\rightarrow Cl(g)+3\:F(g)$

As there are three $Cl-F$ bond,

Therefore, average Cl - F bond energy $= \dfrac{514}{3} = 171$


Hence, option C is correct.

$AB,\, A _2$ and $B _2$ are diatomic molecules. If the bond enthalpies of $A _2,\, AB\, &amp;\, B _2$ are in the ratio 1 : 1 : 0.5 and enthalpy of formation of AB from $A _2$ and $B _2$ is - 100 kJ/mol$^{-1}$. What is the bond enthalpy of $A _2$.

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

  1. 400 kJ/mole

  2. 300 kJ/mole

  3. 500 kJ/mole

  4. 900 kJ/mole

Show answer
Correct Option: A
Explanation:

$\frac {1}{2} A _2 + \frac {1}{2}B _2 \rightarrow AB \Delta H= -100J$

Let the bond enthalpies of $A _2 : B _2 : AB = 1: 0.5 : 1 $
From reaction : $0.5x + 0.25x - 1x = 100 \Rightarrow x= \frac {100}{0.25}= 400kJ $
Bond enthalpy of $A _2= 1x = 400kJmol^{-1}$

From the following thermochemical equations:
$C _2H _4 \rightarrow C _2H _6;     \triangle H = -32.7 kcal$
$C _6H _6+3H _2 \rightarrow C _6H _{12};   \triangle H= -49.2kcal$
Calculate resonance energy of benzene.

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

  1. $-48.9\ kcal$

  2. $-49.2\ kcal$

  3. $-16.5\ kcal$

  4. $-98.1\ kcal$

Show answer
Correct Option: A
Explanation:

Enthalpy of hydrogenation of $C=C$ bond $=-32.7\ kcal$

Calculated  enthalpy of hydrogenation of benzene $=-32.7\times3=-98.1\ kcal$

Actual enthalpy of hydrogenation of benzene$=-45.2\ kcal$
Hence, Resonance energy of benzene$=-98.1-(-45.2)=52.9\ kcal$

Energy required to dissociate $4g$ of gaseous hydrogen into free gaseous atoms is $208\ Kcal$ at ${25}^{o}C$. The bond energy of $H-H$ bond will be:

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

  1. $1.04\ Kcal$

  2. $10.4\ Kcal$`

  3. $104\ Kcal$

  4. $1040\ Kcal$

Show answer
Correct Option: C
Explanation:

The reaction involved is,
$H _2(g) \rightarrow 2H(g)$                


  $\Delta H = \text{H-H bond energy}$

Since, $\text{moles =} \dfrac{\text{mass of compound}}{\text{molecular mass of compound}}$

So, $ \text{moles of} \ H _2 = \dfrac{4}{2} = 2$

Hence, energy required to break 4gm or 2 moles of $H _2$ into gaseous atoms = $208Kcal = 2\times \text{H-H bond energy}$

$\text {H-H bond energy}$ = $104Kcal $ 

Hence, answer is option C.

The dissociation energy of ${CH} _{4}$ is $400kcal$ ${mol}^{-1}$ and that of ethane is $670kcal$ ${mol}^{-1}$. The C-C bond energy is:

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

  1. $270kcal$

  2. $70kcal$

  3. $200kcal$

  4. $240kcal$

Show answer
Correct Option: B
Explanation:

Dissociation reaction of $CH _4$ is,
$CH _4(g) \rightarrow C(g) + 4H(g)$         =>    $\Delta H = 400\text{ Kcal mol}^{-1} = 4\times \text{(bond energy of C-H bond)}$
=> $\text{bond energy of C-H bond = 100Kcal}$
Similarly, dissociation reaction of $C _2H _6$ is,
$C _2H _6(g) \rightarrow 2C(g) +6H(g)$     =>    $\Delta H = 670\text{Kcal mol}^{-1} =\text{(C-C bond energy)} + 6\times \text{(C-H bond energy)} $
=> $\text{(C-C bond energy)} = 670 - 6\times 100 = 70Kcal$ 
Hence, answer is option B.