QJEEM21Ch48aa: JEE Main 2021 Chemistry - Class 11 JEEM21Ch48aa Solution: The ionization enthalpy of Na + formation from Na (g) is 495.8…

Question

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The ionization enthalpy of Na⁺ formation from Na_(g) is 495.8 kJ mol^{$-$ 1}, while the electron gain enthalpy of Br is

-

325.0 kJ mol^{$-$ 1}. Given the lattice enthalpy of NaBr is

-

728.4 kJ mol^{$-$ 1}. The energy for the formation of NaBr ionic solid is (

-

) ____________

\times

10^{$-$ 1} kJ mol^{$-$ 1}.

Accepted Answer

N a (s) \overset{}{⟶} N a^{+} (g)

\Delta H = 495.8

\frac{1}{2} B r_{2} (l) + e^{-} \overset{}{⟶} B r^{-} (g)

\Delta H = 325

N a^{+} (g) + B r^{-} (g) \overset{}{⟶} N a B r (s)

\Delta H = - 728.4

N a (s) + \frac{1}{2} B r_{2} (l) \overset{}{⟶} N a B r (s) .

\Delta H = ?

\Delta H = 495.8 - 325 - 728.4 - 557.6

kJ

= - 5576 \times {10^{ - 1}}

kJ

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