You visited us 1 times! Enjoying our articles? Unlock Full Access!

Heat of Formation

From the foll...

Question

From the following data :
Enthalpy of formation of $C H_{3} C N = 87.86$ kJ / mol
Enthalpy of formation of $C_{2} H_{6} = - 83.68$ kJ / mol
Enthalpy of sublimation of graphite $= 719.65$ kJ / mol
Enthalpy of dissociation of nitrogen $= 945.58$ kJ / mol
Enthalpy of dissociation of $H_{2} = 435.14$ kJ / mol
$C - H$ bond enthalpy $= 414.22$ kJ / mol
Calculate the bond enthalpy of ( i ) $C - C$ ; ( ii ) $C \equiv N$

( i ) 370.58 kJ /

m o l^{- 1}

; ( ii ) 850.2 kJ /

m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

( i ) 343.58 kJ /

m o l^{- 1}

; ( ii ) 891.2 kJ /

m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

( i ) 370.58 kJ /

m o l^{- 1}

; ( ii ) 891.2 kJ /

m o l^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

( i ) 343.58 kJ /

m o l^{- 1}

; ( ii ) 850.2 kJ /

m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C ( i ) 370.58 kJ / $m o l^{- 1}$ ; ( ii ) 891.2 kJ / $m o l^{- 1}$
Bond enthalpy $B E_{C - C} = ?$ $B E_{C \equiv N} = ?$
$(i) B E_{C - C}$

$C_{2} H_{6} (g) \to 2 C$ (graphite) $+ 3 H_{2} (g)$ $Δ_{f} H_{1} = 83.68 k J / m o l$

$2 C$ (graphite) $\to 2 C (g)$ $Δ_{f} H_{2} = 2 \times 719.65 k J / m o l$

$3 H_{2} (g) \to 6 H (g)$ $Δ_{f} H_{3} = 3 \times 435.14 k J / m o l$

Adding we get,
$C_{2} H_{6} (g) \to 2 C (g) + 6 H (g)$ $Δ_{f} H_{4} = 2828.4 k J / m o l$

$Δ_{f} H_{4} = {B E}_{C - C} + 6 {B E}_{C - H}$

so ${B E}_{C - C} = [2828.4 - 6 \times 414.22] = 343.08 k J / m o l$

$(i i) B E_{C \equiv N}$

${C H}_{3} C N (g) \to 2 C$ (graphite) $+ \frac{3}{2} H_{2} (g) + \frac{1}{2} N_{2} (g)$ $Δ_{f} H_{1} = - 87.66 k J / m o l$

$2 C$ (graphite) $\to 2 C (g)$ $Δ_{f} H_{2} = 2 \times 719.65 k J / m o l$

$\frac{3}{2} H_{2} (g) \to 3 H (g)$ $Δ_{f} H_{3} = \frac{3}{2} \times 435.14 k J / m o l$

$\frac{1}{2} N_{2} (g) \to N (g)$ $Δ_{f} H_{4} = \frac{1}{2} \times 945.58 k J / m o l$

Adding we get,
$C H_{3} C N (g) \to 2 C (g) + 3 H (g) + N (g)$ $Δ_{f} H_{5} = 2476.9 k J / m o l$

$Δ_{f} H_{5} = {B E}_{C - C} + 3 {B E}_{C - H} + {B E}_{C \equiv N}$
So ${B E}_{C \equiv N} = (2476.9 - 3 \times 414.22 - 343.08) = 891.2 k J / m o l$

Suggest Corrections

Similar questions

Q. Enthalpy of formation of

C_{2} H_{6} = - 83 k J / m o l

Enthalpy of sublimation of graphite =

719 k J / m o l

Enthalpy of bond dissociation of

H_{2} = 435 k J / m o l

C - H bond enthalpy =

414 k J / m o l

Calculate the bond enthalpy of C - C

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, $- 890.3 k J m o l^{- 1} - 393.5 k J m o l^{- 1}$ , and $- 285.8 k J m o l^{- 1}$ respectively. Enthalpy of formation of $C H_{4 (g)}$ will be
(i) $- 74.8 k J m o l^{- 1}$
(ii) $- 52.27 k J m o l^{- 1}$
(iii) $+ 74.8 k J m o l^{- 1}$
(iv) $+ 52.26 k J m o l^{- 1}$

Q. Bond dissociation enthalpies of

H_{2} (g) a n d N_{2} (g)

are

436.0 k J m o l^{- 1}

and

941.3 k J m o l^{- 1}

respectively. Enthalpy of formation of

N H_{3}

(g) is

- 46.0 k J m o l^{- 1}

. What is the enthalpy of atomization of

N H_{3}

Q. Use the following data to calculate the lattice energy of caesium oxide.
Enthalpy of formation of caesium oxide =

- 233 k J m o l^{- 1}

Enthalpy of sublimation of

C s = + 78 k J m o l^{- 1}

First ionization energy of

C s = + 375 k J m o l^{- 1}

Enthalpy of dissociation of

O_{2} (g) = 494 k J m o l^{- 1} o f O_{2} m o l e c u l e s

First electron affinity of

O = - 141 k J m o l^{- 1} o f O a t o m s

Second electron affinity of

O = + 845 k J m o l^{- 1} o f O^{1 -} i o n s

The standard enthalpy of formation of $N H_{3}$ is $46.0 k J m o l^{- 1}$ . If the enthalpy of formation of $H_{2}$ from its atom is $- 436 k J m o l^{- 1}$ and that of $N_{2}$ is $- 712 k J m o l^{- 1}$ , the average bond enthalpy of $N - H$ bond in $N H_{3}$ is

Join BYJU'S Learning Program