Tag: energetics and thermochemistry

Quizzes (6) Questions (104)

Questions Related to energetics and thermochemistry

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

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 :

  1. 2940 kcal $ mol^{-1} $
  2. 350 kJ $ mol^{-1} $
  3. 700 kJ $ mol^{-1} $
  4. 1470 kcal $ mol^{-1} $
Reveal answer Fill a bubble to check yourself
B,D Correct answer
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$mol^{-1}$

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

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:

  1. $5.0\ \times \ { 10 }^{ -5 }$
  2. $5.0\ \times \ { 10 }^{ 15 }$
  3. $5.0\ \times \ { 10 }^{ -15 }$
  4. $0.2\ \times \ { 10 }^{ 5 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $K$

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

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

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

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

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

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?

  1. $140.9\ kcal$
  2. $49\ kcal$
  3. $40\ kcal$
  4. $76\ kcal$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

C2H4 formation: 2C(s) + 2H2(g) → C2H4; ΔfH = 12.5 kcal. Atomization: 2C(s) → 2C(g); ΔH = 2(171) = 342 kcal. 2H2(g) → 4H(g); ΔH = 2(104.3) = 208.6 kcal. Total atomization = 550.6 kcal. This forms 4 C-H bonds (4×99.3 = 397.2) and 1 C=C bond. BE(C=C) = 550.6 - 397.2 - 12.5 = 140.9 kcal.

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 : $431.37\ kJ\ mol^{-1}$

$C = C$ 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?
  1. $+553.0\ kJ\ mol^{-1}$
  2. $+1523.6\ kJ\ mol^{-1}$
  3. $-243.6\ kJ\ mol^{-1}$
  4. $-120.0\ kJ\ mol^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Delta H = Sum(BE reactants) - Sum(BE products). Reactants: 1(C=C) + 4(C-H) + 1(H-H) = 606.1 + 4(410.5) + 431.37 = 606.1 + 1642 + 431.37 = 2679.47. Products: 1(C-C) + 6(C-H) = 336.49 + 6(410.5) = 336.49 + 2463 = 2799.49. Delta H = 2679.47 - 2799.49 = -120.02 kJ/mol.

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

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?

  1. 463.15 kJ $mol^{-1}$
  2. 428.5 kJ $mol^{-1}$
  3. 69.3 kJ $mol^{-1}$
  4. 926.3 kJ $mol^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $mol^{-1}$

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

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

  1. $-106\ kJ\ mol^{-1}$
  2. $0.632\ kJ\ mol^{-1}$
  3. $60.32\ kJ\ mol^{-1}$
  4. $-10.632\ kJ\ mol^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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

The Van't Hoff equation is :

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $-\Delta  G^{\circ} = RT    log _e   K _p$.

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

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

  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. $T=\frac{\Delta G^{0}}{\Delta H^{0}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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