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Standard XII

Chemistry

Electron Gain Enthalpy

Calculate the...

Question

Calculate the lattice energy of a salt $M X_{(s)}$ from the date given below:
Heat of formation of $M X$ $(Δ H)$ $= - 550$ kJ/mol
Heat of sublimation of $M (S) = 80$ kJ/mol
Heat of dissociation of $X_{2}$ $(D) = 155$ kJ/mol
Ionization energy of $M (I E) = 347$ kJ/mol
Electron affinity of $X (E A) = - 343$ kJ/mol

- 838.5

kJ/mol

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- 938.5

kJ/mol

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- 711.5

kJ/mol

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- 638.5

kJ/mol

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Solution

The correct option is C $- 711.5$ kJ/mol
$Δ H_{f} = S + 0.5 D + I E + E A + U$
$- 550 = 80 + \frac{155}{2} + 347 - 343 + U$
$U = - 711.5 k J / m o l$
Thus, the lattice energy of the salt is $- 711.5$ kJ/mol.

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Similar questions

Q. Calculate the lattice energy of a salt

M X_{(s)}

from the data given below:
Heat of formation of

M X (Δ H) = - 550 kJ/mol

Heat of sublimation of

M (S) = 80 kJ/mol

Heat of dissociation of

X_{2} (D) = 155 kJ/mol

Ionization energy of

M (I) = 347 kJ/mol

Electron affinity of

X (E) = - 343 kJ/mol

Q. Calculate the lattice energy of a salt

M X_{(s)}

from the data given below:
Heat of formation of

M X (Δ H) = - 550 kJ/mol

Heat of sublimation of

M (S) = 80 kJ/mol

Heat of dissociation of

X_{2} (D) = 155 kJ/mol

Ionization energy of

M (I) = 347 kJ/mol

Electron affinity of

X (E) = - 343 kJ/mol

Q. Use the following data to calculate second electron affinity of oxygen for the process

O^{-} (g) + e^{-} (g) \to O^{2 -} (g)

.
Heat of sublimation of

M g (s) = 147.7 k J / m o l

Ionisation energy of

M g (g)

to form

M g^{2 +} (g) = 2189 k J / m o l

Bond dissociation energy for

O_{2}

= 498.4 k J / m o l

First electron affinity of

O (g) = - 141 k J / m o l

Heat formation of

M g O (s) = - 601.7 k J / m o l

lattice energy of

M g O = - 3791 k J / m o 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

Q. Given sublimation and ionization energy of Na are 108 kJ/mol and 496 kJ/mol respectively and bond dissociation energy required to make one mole chlorine atoms and its electron affinity energy are 122 kJ/mol and -349 kJ/mol.If

△ H_{f}^{0}

of NaCl(s) is -411 kJ/mol.What is its approximate lattice formation enthalpy?

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Calculate the lattice energy of a salt MX(s) from the date given below:Heat of formation of MX(ΔH) =−550 kJ/molHeat of sublimation of M(S)=80 kJ/molHeat of dissociation of X2(D)=155 kJ/molIonization energy of M(IE)=347 kJ/molElectron affinity of X(EA)=−343 kJ/mol

The correct option is C −711.5 kJ/molΔHf=S+0.5D+IE+EA+U−550=80+1552+347−343+UU=−711.5kJ/molThus, the lattice energy of the salt is −711.5 kJ/mol.

The correct option is C $- 711.5$ kJ/mol
$Δ H_{f} = S + 0.5 D + I E + E A + U$
$- 550 = 80 + \frac{155}{2} + 347 - 343 + U$
$U = - 711.5 k J / m o l$
Thus, the lattice energy of the salt is $- 711.5$ kJ/mol.