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Electronegativity and Electron Gain Enthalpy

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 Br 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 from Na(g) and Br(g) is (-)__________ $\times 10^{- 1} k J m o l^{- 1}$

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Solution

By Hess law, when there is more that one reaction, the total enthalphy change is the sum of the enthalpy changes of each reaction.

IE - Ionisation enthalpy
EGE - Electron gain enthalpy
The enthalpy of reaction of $N a B r = I E + E G E + Lattice energy$
$Δ_{r} H = 495.8 - 325 - 728.4$
$= - 557.6 k J / m o l$
= $- 5576 \times 10^{- 1} k J / m o l$

Hence, answer is 5576.

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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}

K C l

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

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

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

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

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

K C l

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

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

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

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

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

N H_{3}

- 46.0

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

H_{2}

- 436

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

N_{2}

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

N - H

N H_{3}

C a C l_{2}

C a (s)

121 k J m o l^{- 1}

C l_{2} (g)

2 C l (g)

242.8 k J m o l^{- 1}

C a (g)

C a^{2 +} (g)

2422 k J m o l^{- 1}

C l (g)

C l^{- 1} (g)

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

△_{f} H

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

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