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Covalent Bond vs Ionic Bond

The ionizatio...

Question

The ionization enthalpy of $N a^{+}$ formation from $N a (g)$ is $495.8 k J m o l^{- 1}$ while the electron gain enthalpy of $B r$ is $- 325.0 k J m o l^{- 1}$ . Given the lattice enthalpy of $N a B r$ is $- 728.4 k J m o l^{- 1}$ . The energy for the formation of $N a B r$ ionic solid is (–) _____ $\times 10^{- 1} k J m o l^{- 1}$ .

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Solution

$N a (s) \to N a^{+} (g) + e^{-} Δ H = 495.8 k J m o l^{- 1}$

$\frac{1}{2} B r_{2} (l) + e^{-} \to B r^{-} (g) Δ H = - 325.0 k J m o l^{- 1}$

$N a^{+} (g) + B r^{-} (g) \to N a B r (s) Δ H = - 728.4 k J m o l^{- 1}$

On adding the above three equations,
$N a (s) + \frac{1}{2} B r_{2} (l) \to N a B r (s) Δ H_{f} = ?$
$Δ H_{f} = 495.8 - 325.0 - 728.4 = - 557.6 k J m o l^{- 1} Δ H_{f} = - 5576 \times 10^{- 1} k J m o l^{- 1}$

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Similar questions

N a^{+}

N a (g)

495.8 k J m o l^{- 1}

B r

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

N a B r

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

N a B r

\times 10^{- 1} k J m o l^{- 1}

N a^{+}

N a (g)

495.8 k J m o l^{- 1}

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

N a B r

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

N a B r

\times 10^{- 1} k J m o l^{- 1}

C a C l_{2}

C a (s) \to C a (g) = 121 k J m o l^{- 1}

C l_{2} (g) \to 2 C l (g) = 242.8 k J m o l^{- 1}

C a (g) \to C a^{+ +} (g) = 2422 k J m o l^{- 1}

2 C l \to 2 C l^{- 1} = (2 \times - 355) k J m o l^{- 1} = - 710 k J m o l^{- 1}

C a C l_{2} = - 795 k J m o l^{- 1}

{kJ mol}^{- 1}

L i F

155.3 {kJ mol}^{- 1}

F_{2} = 75.3 kJ

= 520 {kJ mol}^{- 1}

F (g) = - 333 kJ Δ_{f} H_{overall} = - 594 {kJ mol}^{- 1}

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

C s = + 78 k J m o l^{- 1}

C s = + 375 k J m o l^{- 1}

O_{2} (g) = 494 k J m o l^{- 1} o f O_{2} m o l e c u l e s

O = - 141 k J m o l^{- 1} o f O a t o m s

O = + 845 k J m o l^{- 1} o f O^{1 -} i o n s

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