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

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

  1. $390.1 kJ/mol$
  2. $103.9 kJ/mol$
  3. $903.1 kJ/mol$
  4. $309.1 kJ/mol$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $S-F = 1855/6 = 309.1 kJ/mol$
Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

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

  1. $-433 kJ$
  2. $+433 kJ$
  3. $-216 kJ$
  4. $+216 kJ$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 ${ H } _{ 2 }=\dfrac { 433 }{ 2 } =+216.5kJ$

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

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] :

  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 $
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $mol^{-1}$