JEE PYQ: Chemical Equilibrium Question 9
Question 9 - 2021 (25 Feb Shift 1)
The ionization enthalpy of $\text{Na}^+$ formation from $\text{Na}{(g)}$ is $495.8 \text{ kJ mol}^{-1}$, while the electron gain enthalpy of Br is $-325.0 \text{ kJ mol}^{-1}$. Given the lattice enthalpy of NaBr is $-728.4 \text{ kJ mol}^{-1}$. The energy for the formation of NaBr ionic solid is $(-) \text{___} \times 10^{-1} \text{ kJ mol}^{-1}$
Show Answer
Answer: 5576
Solution
$\Delta H = 495.8 - 325 - 728.4 = -557.6 \text{ kJ} = -5576 \times 10^{-1} \text{ kJ}$.
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