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Standard XII

Chemistry

Heat of Formation

The standard ...

Question

The standard enthalpy of formation $(Δ_{f} H_{298}^{o})$ for methane, $C H_{4}$ is $- 74.9 k J m o l^{- 1}$ . In order to calculate the average energy given out in the formation of a $C - H$ bond from this, it is necessary to know which one of the following?

The dissociation energy of the hydrogen molecule,

H_{2}

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The first four ionisation energies of carbon.

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The dissociation energy of

H_{2}

and enthalpy of sublimation of carbon (graphite).

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The first four ionisation energies of carbon and electron affinity of hydrogen.

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Solution

The correct option is C The dissociation energy of $H_{2}$ and enthalpy of sublimation of carbon (graphite).
$C_{(s, g r a p h i t e)} + 2 H_{2} \to C H_{4}$
In order to calculate the average energy given out in the formation of a $C - H$ bond from this, it is necessary to know the dissociation energy of $H_{2}$ and enthalpy of sublimation of carbon (graphite).

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Similar questions

Q. Calculate the lattice enthalpy of

K C l

from the following data at standard states-
Enthalpy of sublimation of

K = 89 k J m o l^{- 1}

Enthalpy of dissociation of

C l_{2} = 244 k J m o l^{- 1}

Ionisation energy of

K = 425 k J m o l^{- 1}

Electron gain enthalpy of

C l = - 355 k J m o l^{- 1}

Enthalpy of formation of

K C l = - 438 k J m o l^{- 1}

Q. Cesium chlorides is formed according to the following equation

C s (s) + 0.5 C l_{2} (g) \to C s C l (s)

. The enthalpy of sublimation of Cs, the enthalpy of dissociation of chlorine, ionization energy of

C s

and electron affinity of cholrine are

81.2, 243.0, 375.7

and

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

. The energy involved in the formation of

C s C l

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

. Calculate the lattice energy of

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Q. Caesium chlorides is formed according to the following equation

C s (s) + 0.5 C l_{2} (g) \to C s C k (s)

. The enthalpy of sublimation of

C s

, enthalpy of dissociation of chlorine, ionization energy of

C s

and electron affinity of chlorine are

81.2, 243.0, 375.7

and

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

. The energy change involved in the formation of

C s C l

- 388.6 K J m o l^{- 1}

. Calculate the lattice energy of

C s C l

Q. Caesium chloride is formed according to the following equation:

C s (s) + 0.5 {C l}_{2} (g) ⟶ C s C l (s)

The enthalpy of sublimation of

C s

, enthalpy of dissociation of chlorine, ionization energy of

C s

and electron affinity of chlorine are

81.2, 243.0, 375.7

and

- 348.3 k J

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C s C l

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

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C s C l

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

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The standard enthalpy of formation (ΔfHo298) for methane, CH4 is −74.9kJmol−1. In order to calculate the average energy given out in the formation of a C−H bond from this, it is necessary to know which one of the following?

The standard enthalpy of formation $(Δ_{f} H_{298}^{o})$ for methane, $C H_{4}$ is $- 74.9 k J m o l^{- 1}$ . In order to calculate the average energy given out in the formation of a $C - H$ bond from this, it is necessary to know which one of the following?