Calculate Hor for the reaction CH2Cl2g - Cg - 2Hg - 2Clg-The average bond enthalpies of C - H and C - Cl bonds are 414 kJ mol-1 and 330 kJ mol-1-

We are given nbsp;ε _C H = nbsp;414 kJ mol^ 1 nbsp; ε _C Cl = 330 kJ mol^ 1 We know Δ _rH= ∑ (Bond Energy)_reactants ∑ (Bond Energy)_products Here th ...

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

Byju's Answer

Standard XII

Chemistry

Heat of Reaction

Calculate Hor...

Question

Calculate $△ H_{r}^{o}$ for the reaction $C H_{2} C l_{2} (g) \to C (g) + 2 H (g) + 2 C l (g) .$
The average bond enthalpies of C - H and C - Cl bonds are $414 k J m o l^{- 1}$ and $330 k J m o l^{- 1} .$

1488 k J m o l^{- 1}

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

1468 k J m o l^{- 1}

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

1478 k J m o l^{- 1}

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

1498 k J 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 A $1488 k J m o l^{- 1}$
We are given
$ε_{C - H} = 414 k J m o l^{- 1}$
$ε_{C - C l} = 330 k J m o l^{- 1}$

We know
$Δ_{r} H = \sum (B o n d E n e r g y)_{r e a c t a n t s} - \sum (B o n d E n e r g y)_{p r o d u c t s}$
Here the bond energy of products won't be taken into account as the products doesn't contain any bonds.
So,
$△ H_{r} = 2 ε_{C - H} + 2 ε_{C - C l}$
putting the values,
$= 2 \times 414 + 2 \times 330$
$= 828 + 660 = 1488 k J / m o l .$
hence, $△ H_{r} = 1488 k J / m o l .$

Theory:

Determination of enthalpy of reaction:
$Δ_{r} H = \sum (B o n d E n e r g y)_{r e a c t a n t s} - \sum (B o n d E n e r g y)_{p r o d u c t s}$

Suggest Corrections

Similar questions

Q. Calculate

△ H_{r}^{o}

for the reaction

C H_{2} C l_{2} (g) \to C (g) + 2 H (g) + 2 C l (g) .

The average bond enthalpies of C - H and C - Cl bonds are

414 k J m o l^{- 1}

and

330 k J m o l^{- 1} .

Q. If enthalpy of a reaction is

1482 k J / m o l

and bond dissociation enthalpy of

(C - H)

bond is

415 k J / m o l

than bond dissociation enthalpy of

(C - C l)

bond is:

C H_{2} C l_{2 (g)} \to C_{(g)} + 2 H_{(g)} + 2 C l_{(g)}

Q. Calculate the enthalpy change

(△ H)

of the following reaction

2 C_{2} H_{2} (g) + 5 O_{2} (g) \to 4 C O_{2} (g) + 2 H_{2} O (g)

given average bond enthalpies of various bonds, i.e.,

C - H, C \equiv C, O = O, C = O, O - H

as 414, 814 499, 724 and 640

k J m o l^{- 1}

respectively.

Q. Calculate

C - C l

bond enthalpy from following reaction :

C H_{3} C l (g) + {C l}_{2} (g) ⟶ C H_{2} {C l}_{2} (g) + H C l (g)

Δ H^{o} = - 104 k J

. If

C - H

C l - C l

and

H - C l

bond enthalpies are

414

243

and

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

respectively.

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

Join BYJU'S Learning Program

Explore more

Heat of Reaction

Standard XII Chemistry

Join BYJU'S Learning Program

Calculate △Hor for the reaction CH2Cl2(g)→C(g)+2H(g)+2Cl(g). The average bond enthalpies of C - H and C - Cl bonds are 414 kJmol−1 and 330 kJ mol−1.

Calculate $△ H_{r}^{o}$ for the reaction $C H_{2} C l_{2} (g) \to C (g) + 2 H (g) + 2 C l (g) .$
The average bond enthalpies of C - H and C - Cl bonds are $414 k J m o l^{- 1}$ and $330 k J m o l^{- 1} .$