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Byju's Answer

Standard XII

Physics

Potential Energy of a Dipole in Uniform Electric Field

The correct f...

Question

The correct folrmula of Q by using Born-Haber cycle for the preparation of $M g B r_{2} (s)$ will be:
S $=$ heat of sublimation of Mg(s)
I $=$ ionisation energy of Mg(g)
D $=$ dissociation energy of Br $_{2}$ (g)
-E $=$ electron affinity of Br(g)
-U $=$ lattice energy of MgBr $_{2}$ (s)
Q $=$ heat of formation of MgBr $_{2}$ (s)

Q = S + I + D - 2 E - U

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Q = S + I + D - E - 2 U

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Q = S - I + D - 2 E - U

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None of these

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Solution

The correct option is A $Q = S + I + D - 2 E - U$
$Δ H_{Π} = S + I + D - 2 E - U = Δ H_{1} = Q$
S $=$ heat of sublimation of Mg(s)
I $=$ ionisation energy of Mg(g)
D $=$ dissociation energy of Br $_{2}$ (g)
-E $=$ electron affinity of Br(g)
-U $=$ lattice energy of MgBr $_{2}$ (s)
Q $=$ heat of formation of MgBr $_{2}$ (s)

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Similar questions

The lattice energy of NaCl(s) using the following data will be:

heat of sublimation of $N a (s) = S$

$(I E)_{1}$ of $N a (g) = I$

bond dissociation energy of $C l_{2} (g) = D$

electron affinity of $C l (g) = - E$

heat of formation of

N a C l (s) = - Q

Q. According to Born-Haber's cycle, the enthalpy of formation of ionic compounds can be determined. The formation of NaCl involves following steps –
(i)

N a (s) S - ------------ \to Energy of subtimation N a (g)

(ii)

N a (g) I - ---------- \to Ionisation potenital N a^{+} (g) + e^{-}

(iii)

C l_{2} (g) D - -------------- \to Bond dissociation energy 2 C l (g)

(iv)

C l (g) + e^{-} E - --------- \to Electron - affinity C l^{-} (g)

(v)

C l^{-} (g) + N a^{+} (g) U - ------- \to Lattice energy N a C l (s)

The enthalpy of formation of NaCl will be -

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 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 kJ/mol

Ionisation energy of

M g (g)

to form

M g^{2 +} (g) = 2189 kJ/mol

Bond dissociation energy for

O_{2}

= 498.4 kJ/mol

First electron affinity of

O (g) = - 141 kJ/mol

Heat formation of

M g O (s) = - 601.7 kJ/mol

lattice energy of

M g O = - 3791 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

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The correct folrmula of Q by using Born-Haber cycle for the preparation of MgBr2(s) will be:S = heat of sublimation of Mg(s)I = ionisation energy of Mg(g)D = dissociation energy of Br2(g)-E = electron affinity of Br(g)-U = lattice energy of MgBr2(s)Q = heat of formation of MgBr2(s)

The correct option is A Q=S+I+D−2E−UΔHΠ=S+I+D−2E−U=ΔH1=QS = heat of sublimation of Mg(s)I = ionisation energy of Mg(g)D = dissociation energy of Br2(g)-E = electron affinity of Br(g)-U = lattice energy of MgBr2(s)Q = heat of formation of MgBr2(s)