Tag: energetics and thermochemistry

Questions Related to energetics and thermochemistry

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} $

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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 $mol^{-1}$

At equilibrium, the value of equilibrium constant $K$ is:

chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

  1. $1$

  2. $2$

  3. $3$

  4. $0$

Show answer
Correct Option: A
Explanation:

At equilibrium $\Delta G=0$
$\Delta G=-nRT ln K$
$0=-nRT ln K$
or $K=1$

The equilibrium constants of a reaction is $73$. Calculate standard free energy change.

chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

  1. $-106\ kJ\ mol^{-1}$

  2. $0.632\ kJ\ mol^{-1}$

  3. $60.32\ kJ\ mol^{-1}$

  4. $-10.632\ kJ\ mol^{-1}$

Show answer
Correct Option: D
Explanation:

${ \triangle G }^{ 0 }=-RTlnK$             $T={ 27 }^{ 0 }C=300K$

${ \triangle G }^{ 0 }=-8.3\times 300\times ln73$
          $=-10.632KJ{ mol }^{ -1 }$

The Van't Hoff equation is :

chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

  1. $\Delta G^{\circ} = RT log _e K _p$

  2. $-\Delta G^{\circ} = RT log _e K _p$

  3. $\Delta G^{\circ} = RT^2 lnK _p$

  4. None of the above

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Correct Option: B
Explanation:

The Van't Hoff equation gives the relationship between the standard gibbs free energy change and the equilibrium constant.  It is represented by the equation $-\Delta  G^{\circ} = RT    log _e   K _p$.

Standard Gibbs Free energy change $\Delta { G }^{ o }$ for a reaction is zero. The value of equilibrium constant of the reaction will be:

chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

  1. 0

  2. 1

  3. 2

  4. 3

Show answer
Correct Option: B
Explanation:

$\text{Option B is correct.}$


$\Delta{G^o}=-RTlog K _c$
$0=-nRTlogK _c$
$logK _c=0$
$K _c=1$

if for the heterogeneous equilibrium $CaCO _{3}(s)\rightleftharpoons CaO(s)+CO _{2}(g);$ K=1 at 1 atm, the temperature is given by:

chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

  1. $T=\frac{\Delta S^{0}}{\Delta H^{0}}$

  2. $T=\frac{\Delta H^{0}}{\Delta S^{0}}$

  3. $T=\frac{\Delta G^{0}}{ R^{0}}$

  4. $T=\frac{\Delta G^{0}}{\Delta H^{0}}$

Show answer
Correct Option: B
Explanation:

$\Delta G = 2.303RT\space logK$


As K =1 , $\Delta G = 0$

We know the relation,

$\Delta G = \Delta H - T\Delta S$

$T = \dfrac{\Delta H}{\Delta S}$

Option B is correct