Bond enthalpies - class-XII

The statement about dilution enthalpy is incorrect because enthalpy of dilution DOES depend on the original concentration of the solution. Dilution from different starting concentrations releases different amounts of heat. The other statements are correct: Born-Haber cycle relation, bond enthalpy order (N≡N=946 kJ/mol > C≡N=891 > C≡C=619), and ice melting thermodynamics.

$\Delta { H } _{ reaction }=\sum { { \left( BE \right)  } _{ reactants } } -\sum { { \left( BE \right)  } _{ products } } $
$-90=x+3\times 435-6\times 390$
$x=945kJ$ ${mol}^{-1}$

${ H } _{ reaction }=\sum { { BE } _{ reactants } } -\sum { { BE } _{ products } } $
$-46=(\cfrac(712)+\cfrac{3}{2}(436))-3x$
$x=+352\ kJ{ mol }^{ -1 } $

Using bond enthalpy method: ΔH = Σ(bonds broken) - Σ(bonds formed). Breaking: 4 C-H (4×414=1656), 1 C=C (619), 3 O=O (3×499=1497). Total broken = 3772. Forming: 4 C=O (4×724=2896), 4 O-H (4×460=1840). Total formed = 4736. ΔH = 3772-4736 = -964 kJ. The reaction is exothermic.

$H _2+Cl _2 \longrightarrow 2HCl+44Kcal$

The standard enthalpies of n-pentane, isopentane and neopentane are -35.0, -37.0 and -40.0 K.cal/mole respectively. The most stable isomer of pentane in terms of energy is neopentane as it has most negative value of the standard enthalpy.

Using Born-Haber type approach: ΔfH(PCl3) = ΔHatom(P) + 3ΔHatom(Cl) - 3BE(P-Cl). BE(P-Cl) = [314 + 3(121) - 306]/3 = 123.67 kJ/mol. The calculation uses formation enthalpy as the difference between atomization and bond formation energies.

Lattice enthalpy is inversely proportional to inter-ionic distance. LiF has the smallest ions and thus the highest lattice energy, followed by LiCl, NaF, and NaCl (larger ions).

${ \Delta } _{ f }{ H }^{ o }(S,monolithic)\ne 0$ as it is not elemental form of S.
$\Delta E=0$ for combustion of ${ C } _{ 2 }{ H } _{ 6 }(g)$ in a sealed rigid adiabatic container as in adiabatic process energy exchange is zero.
For ${ C } _{ 2 }{ H } _{ 6 }(g)$,
$BE _{C-C} = 2812 - 6\times BE _{C-H} = 2812-6\times 1656/4 = 328$kJ/mol

The bond energy   of a $C-! ! ! -C$ single bond is approximately 100 $kcal : mol^{-1}$. 

Fluorine is more electron-negative than oxygen and oxygen is more electro-negative than nitrogen.

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

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.

As we know,

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.

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

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

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

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.

$\Delta H $ of the given reaction,
$C(g) + 4H(g) \rightarrow CH _4(g)$ 
can be written as, $\Delta H = \text {heat released in formation of 4 C-H bonds in CH} _4$
=> $\Delta H = -4\times \text{bond dissociation energy of C-H bond in CH} _4$
=> $-397.8 = -4\times \text{bond dissociation energy of C-H bond in CH} _4$
=> $\text{bond dissociation energy of C-H bond in CH} _4 = 99.45Kcal$
Hence, answer is option A.

$H-H\longrightarrow H+H\quad \quad ;\quad \Delta H$

${C} _{2}{H} _{5}-S-{C} _{2}{H} _{5}+S(s)\rightarrow {C} _{2}{H} _{5}-S-S-{C} _{2}{H} _{5}$
$\Delta { H } _{ reaction }=\sum { { \Delta H } _{ f(products) }^{ o } } +\sum { \Delta { H } _{ f(reactants) }^{ o } } $
$=(-202)-(-147)=-55kJ$
$\Delta { H } _{ reaction }=\sum { { BE } _{ reactants } } -\sum { { BE } _{ products } } $
$-55=$ Heat of sublimation or enthalpy of atomisation of sulphur$-BE(S-S)$
$-55=223-BE(S-S)$
$BE(S-S)=223+55=278kJ$

$ \Delta H _R = (bond energy) _R - (bond energy) _P$

$H-H+Cl-Cl\rightarrow 2H-Cl$

$C-E$ bond has highest bond energy; it means that the covalent bond $C-E$ will be strongest. Smaller is the size of atom, stronger is the covalent bond.

Delta H = Energy of bonds broken - Energy of bonds formed. Bonds broken: (CH3)2CH-H (95) + Cl-Cl (58) = 153. Bonds formed: (CH3)2CH-Cl (80) + H-Cl (103) = 183. Delta H = 153 - 183 = -30 kcal/mol.

This is the same question as 454015. Propagation step: (CH3)2CH-H + Cl-Cl → (CH3)2CH-Cl + H-Cl. ΔH = (95 + 58) - (80 + 103) = -30 kcal/mol. The table gives bond dissociation energies for similar compounds to estimate the reaction enthalpy.

Delta H = Sum(BE reactants) - Sum(BE products). Delta H = [BE(H-H) + BE(Br-Br)] - 2*BE(H-Br) = [433 + 192] - 2(364) = 625 - 728 = -103 kJ.

$NO(g)+O _{ 3 }(g)\rightarrow NO _{ 2 }(g)+O _{ 2 }(g)\quad ;\quad \Delta H=-198.9kJ/mol-----(i)\ \quad \quad \quad \quad \quad \quad \quad O _{ 3 }(g)\rightarrow \cfrac { 3 }{ 2 } O _{ 2 }(g)\quad ;\quad \Delta H=-142.3kJ/mol\ \cfrac { 3 }{ 2 } O _{ 2 }(g)\rightarrow O _{ 3 }(g)\quad ;\quad \Delta H=142.3kJ/mol-----(ii)\ \quad \quad \quad \quad \quad \quad \quad \quad O _{ 2 }(g)\rightarrow 2O(g)\quad ;\quad \Delta H=+495.0kJ/mol\ \quad \quad \quad \quad \quad \quad \quad \quad 2O(g)\rightarrow O _{ 2 }(g)\quad ;\quad \Delta H=-495.0kJ/mol\ O(g)\rightarrow \cfrac { 1 }{ 2 } O _{ 2 }(g)\quad ;\quad \Delta H=-\cfrac { 495.0 }{ 2 } kJ/mol-----(iii)\ Adding\quad (i),\quad (ii)\quad and\quad (iii),\ NO(g)+O(g)\rightarrow NO _{ 2 }(g)\quad ;\quad \Delta H=(-198.9+142.3-\cfrac { 495.0 }{ 2 } )kJ/mol\ \therefore \Delta H=-304.1kJ/mol$

The average C-H bond energy relates to atomization of methane. C(g) + 4H(g) → CH4(g) releases X2 kJ (4 bonds form). So average C-H bond energy = X2/4 kJ/mol. X1 is the formation from elements, not atomization from gaseous atoms.

Carbon has all 4 valence electrons. So it easily bonds with other carbon atoms to form long chains Carbon atoms. Apart from single bond carbon also forms double bonds in compounds like alkenes or triple bonds in compounds like alkynes.

A carbon-carbon bond is very strong. The carbon to carbon triple bond has dissociation energy of $348 \frac{kJ}{mol}$, followed by a double bond with $614 \frac{kJ}{mol}$ and single bond having $839 \frac{kJ}{mol}$.

 The correct order of bond energy strength is option D.

$\frac{1}{2} H _2 + \frac{1}{2} Cl _2 \rightarrow HCl$
$\Delta H _f(HCl) = 91 \ kJ \ mol^{-1}$, H - H = 430 kJ $mol^{-1}$, Cl - Cl = 242 kJ $mol^{-1}$
$91 = \frac{1}{2} \times 430 + \frac{1}{2} \times 242 - B.E. (H - Cl)$
B.E. (H-Cl) = 245 kJ

From definition of Heat of Formation and Hess' law, $\Delta H _r^0 = \sum H _{f(products)}^0 - \sum H _{f(reactants)}^0$

$2CI _2 + 2H _2O \rightarrow 4HCI + O _2$   

The energy required to break a specific covalent bond in one mole of gaseous molecules is called the bond energy or the bond dissociation energy or bond enthalpy.

Bond energy is defined as the energy required to break one mole of bonds in the gaseous state. This is equivalent to the energy released when one mole of bonds is formed from gaseous atoms.

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 ${ H } _{ 2 }=\dfrac { 433 }{ 2 } =+216.5kJ$

$ 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$mol^{-1}$

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.

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.

Heat of dissociation $=(55.9-12.1)=43.8\ KJ$