Tag: chemistry

Questions Related to chemistry

The average, $S - F$ bond energy in $SF _6$ if the $\Delta H^{\circ} _f$ value are $-1100, +275$ and $+80 kJ/mol$ respectively for $SF _6(g),$ S(g) and F(g) is

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

  1. $390.1 kJ/mol$

  2. $103.9 kJ/mol$

  3. $903.1 kJ/mol$

  4. $309.1 kJ/mol$

Show answer
Correct Option: D
Explanation:
As we know,
Heat of a reaction = Bond energy of reactants - Bond energy of products
here,
$S(s) +3F _2(g) \rightarrow SF _6(g)$
so heat of reaction = 
$-1100 = 275 + 6*80$ - bond energy of $SF _6$ 
bond energy of $SF _6$ $= 1855$
so bond energy of $S-F = 1855/6 = 309.1 kJ/mol$

Bond energy of hydrogen gas is $-433 kJ$. How much is the bond dissociation energy of $0.5 mole$ of hydrogen gas?

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

  1. $-433 kJ$

  2. $+433 kJ$

  3. $-216 kJ$

  4. $+216 kJ$

Show answer
Correct Option: D
Explanation:

Bond dissociation energy $= -$ bond formation energy
Given, bond energy of hydrogen $= -433 kJ$
$\therefore$ Bond dissociation energy of one mole ${ H } _{ 2 } = 433 kJ$
$\therefore$ Bond dissociation energy of $0.5$ mole ${ H } _{ 2 }=\dfrac { 433 }{ 2 } =+216.5kJ$

A vessel contains 100 litres of a liquid x. Heat is supplied to the liquid in such a fashion that, heat gives a change in enthalpy. The volume of the liquid increases by 2 litres. If the external pressure is one atm, and 202.6 Joules of heat supplied, then [U $\rightarrow$ total internal energy] :

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

  1. $\Delta U=0, \Delta H=0$

  2. $\Delta U=+202.6 J, \Delta H=+202.6 J$

  3. $\Delta U=-202.6 J, \Delta H=-202.6 J$

  4. $\Delta U=0, \Delta H=+202.6 L $

Show answer
Correct Option: D

Enthalpy of polymerisation of ethylene, as represented by the reaction, $ nCH _2 = CH _2 \rightarrow {(-CH _2- CH _2-)} _n   $ is -100kJ per mole of ethylene.Given bond enthalpy of $ C = C $ bond is 600 kJ$ mol^{-1} $ , enthalpy of $ C - C $ bond (in kJ mol) will be :

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

  1. 2940 kcal $ mol^{-1} $

  2. 350 kJ $ mol^{-1} $

  3. 700 kJ $ mol^{-1} $

  4. 1470 kcal $ mol^{-1} $

Show answer
Correct Option: B,D
Explanation:

$ nCH _2 = CH _2 \rightarrow (-CH _2 - CH _2- ) _n    \Delta H $= 100 KJ/mole
Double bond of ethylene converted to two single bonds.
$ \Delta H $ = - 100 =$ B.E. _{C=C} - 2 B.E. _{C-C}$ 
$ \implies $ - 100 = 600 - $ 2 \times B.E. _{C-C} \implies B.E. _{C-C}$ = 350 KJ/mole = 1470kcal$mol^{-1}$

The first and second dissociation constant of an acid ${ H } _{ 2 }A$ are $1.0\ \times \ { 10 }^{ -5 }$ and $5.0\ \times \ { 10 }^{ -10 }$ respectively. The over all dissociation constant of the acid will be:

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

  1. $5.0\ \times \ { 10 }^{ -5 }$

  2. $5.0\ \times \ { 10 }^{ 15 }$

  3. $5.0\ \times \ { 10 }^{ -15 }$

  4. $0.2\ \times \ { 10 }^{ 5 }$

Show answer
Correct Option: C
Explanation:

$H _2A\overset {K _1}{\rightleftharpoons} HA^-+H^+$


$\Rightarrow K _1=\cfrac {[HA^-][H^+]}{[H _2A]}$  $\longrightarrow (1)$


$HA^-\overset {K _2}{\rightleftharpoons} H^++A^{2-}$

$\Rightarrow K _2=\cfrac {[H^+][A^{2-}]}{[HA^-]}$    $\longrightarrow (2)$

Overall dissociation constant $K$

$\Rightarrow K=\cfrac {[H^+]^2[A^{2-}]}{[H _2A]}=K _1\times K _2$

$=1\times 10^{-5}\times 5\times 10^{-10}$

$=5\times 10^{-15}$ .

$\triangle H _{f} (C _{2}H _{4}) = 12.5\ kcal$

Heat of atomisation of $C = 171\ kcal$
Bond energy of $H _{2} = 104.3\ kcal$
Bond energy $C - H = 99.3\ kcal$

What is $C = C$ bond energy?

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

  1. $140.9\ kcal$

  2. $49\ kcal$

  3. $40\ kcal$

  4. $76\ kcal$

Show answer
Correct Option: A
$\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 : $431.37\ kJ\ mol^{-1}$

$C = C$ bond energy : $606.10\ kJ\ mol^{-1}$

$C - C$ bond energy : $336.49\ kJ\ mol^{-1}$

$C - H$ bond energy : $410.50\ kJ\ mol^{-1}$

Enthalpy for the reactions, will be?
chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

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

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

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

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

Show answer
Correct Option: D

Dissociation of water takes place in two steps:
$H _2O \rightarrow H^+ + OH^-$; $\Delta H$ = +497.8 kJ
$OH^- \rightarrow H^+ + O^{2-}$; $\Delta H$ = +428.5 kJ
What is the bond energy of O - H bond?

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

  1. 463.15 kJ $mol^{-1}$

  2. 428.5 kJ $mol^{-1}$

  3. 69.3 kJ $mol^{-1}$

  4. 926.3 kJ $mol^{-1}$

Show answer
Correct Option: A
Explanation:

In given 2 reactions, $\Delta H$ is basically representing bond energies of H-O bond. So, our answer should be average of these two $\Delta H$ values.
Hence, Average of two bond dissociation energies:    $\frac{497.8 + 428.5}{2}$ = 463.15kJ $mol^{-1}$

Boiling point of heavy water is:

chemistry hard water and soft water heavy water study of heavy water hydrogen and its compounds

  1. $3.82:^0C$

  2. $42:^{0}C$

  3. $100:^{0}C$

  4. $101.42:^{0}C$

Show answer
Correct Option: D
Explanation:

The boiling point of heavy water is $\displaystyle 101.42^oC $ which is higher than the boiling point of normal water $\displaystyle (100^oC) $ due to its higher molar mass.

When $SO _{3}$ treated with $D _{2}O$ , the products are:

chemistry hard water and soft water heavy water study of heavy water hydrogen and its compounds

  1. deutero sulphuric acid

  2. deuterium sulphuric acid

  3. deuterium and sulphuric acid

  4. $D _{2}SO _{3}$

Show answer
Correct Option: A
Explanation:
When $SO _3$ treated with $D _2O$ , the product is deutero sulphuric acid.
$SO _3+D _2O\rightarrow D _2SO _4$.