CameraIcon

SearchIcon

MyQuestionIcon

116

You visited us 116 times! Enjoying our articles? Unlock Full Access!

Heat of Formation

What is the h...

Question

What is the heat of formation of liquid methyl alcohol in $k J m o l^{- 1},$ use the following data.
Heat of vaporization of liquid methyl alcohol = $38 k J m o l^{- 1}$ .
Heat of formation of gaseous atoms from the elements in their standard states:
$H = 218 k J m o l^{- 1}, C = 715 k J m o l^{- 1}, O = 249 k J m o l^{- 1}$
Average bond energies :
$C - H = 415 k J m o l^{- 1} C - O = 356 k J m o l^{- 1} O - H = 463 k J m o l^{- 1}$

A

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

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

B

+ 266 k J m o l^{- 1}

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

C

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

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

D

+ 256 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 C $- 266 k J m o l^{- 1}$
We are given:
$C H_{3} O H (g) \to C H_{3} O H (l); Δ H = 38 k J m o l^{- 1}$

$C (g) + 4 H (g) + O (g) \to C H_{3} O H (g);$ $Δ H = - (3 \times 415 + 356 + 463) = - 2064 k J m o l^{- 1}$

$C (g r a p h i t e) \to C (g);$ $Δ H = 715 k J m o l^{- 1}$

$2 H_{2} (g) \to 4 H (g);$
$Δ H = 2 \times 2 \times 218 = 872 k J m o l^{- 1}$

$\frac{1}{2} O_{2} (g) \to O (g);$ $Δ H = 249 k J m o l^{- 1}$

Adding all the above equations we get our desired equation.: $C (g r a p h i t e) + 2 H_{2} (g) + \frac{1}{2} O_{2} (g) \to C H_{3} O H (l)$ $Δ H = S u m o f a l l = Δ H^{'} s - 266 k J m o l^{- 1}$

Suggest Corrections

thumbs-up

5

Similar questions

Q. What is the heat of formation of liquid methyl alcohol in

k J m o l^{- 1},

use the following data.
Heat of vaporization of liquid methyl alcohol =

38 k J m o l^{- 1}

.
Heat of formation of gaseous atoms from the elements in their standard states:

H = 218 k J m o l^{- 1}, C = 715 k J m o l^{- 1}, O = 249 k J m o l^{- 1}

Average bond energies :

C - H = 415 k J m o l^{- 1} C - O = 356 k J m o l^{- 1} O - H = 463 k J m o l^{- 1}

Q. Compute the heat of formation of liquid methyl alcohol (in kJ mol

^{- 1}

) using the following data. The heat of vaporisation of liquid methyl alcohol is 38 kJ mol

^{- 1}

The heat of formation of gaseous atoms from the elements in their standard states:

H =

^{- 1}

C = 715

^{- 1}

O = 249

^{- 1}

Average bond energies:

C - H = 415

^{- 1}

C - O = 356

^{- 1}

O - H = 463

^{- 1}

Q. Compute the heat of formation of liquid methyl alcohol in kilojoules per mole using the following data.
The heat of vaporisation of liquid methyl alcohol =

38 k J / m o l

.
The heats of formation of gaseous atoms from the elements is their standard states:

H, 218 k J / m o l

C, 713 k J / m o l

O, 249 k J / m o l

.
Average bond energies:

C - H = 415 k J / m o l

C - O = 365 k J / m o l

O - H = 463 k J / m o l

.

Q. Compute the heat of formation of liquid methyl alcohol in kilojoules per mole using the following data.
The heat of vaporisation of liquid methyl alcohol =

38 k J / m o l

.
The heats of formation of gaseous atoms from the elements in their standard states:

H, 218 k J / m o l

C, 715 k J / m o l

O, 249 k J / m o l

.
Average bond energies:

C - H = 415 k J / m o l

C - O = 365 k J / m o l

O - H = 463 k J / m o l

.

Q. Compute the enthalpy of formation of liquid methyl alcohol in kJ

{m o l}^{- 1}

, using the following data. Enthalpy of vaporization of liquid

{C H}_{3} O H = 38 k J / m o l

.
Enthalphy of formation of gaseous atoms from the elements in their standard states are enthalpy of formation of gaseous atoms from the elements in their standard states are

H

\to

C

\to

O

\to

249 kJ / mol
Bond Enthalpies

C - H

\to

C - O

\to

O - H

\to

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Heat of Formation

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon