JEE PYQ: Chemical Equilibrium Question 14
Question 14 - 2020 (05 Sep Shift 2)
Lattice enthalpy and enthalpy of solution of NaCl are 788 kJ mol$^{-1}$ and 4 kJ mol$^{-1}$, respectively. The hydration enthalpy of NaCl is:
(1) $-780$ kJ mol$^{-1}$
(2) $780$ kJ mol$^{-1}$
(3) $-784$ kJ mol$^{-1}$
(4) $784$ kJ mol$^{-1}$
Show Answer
Answer: (3) $-784$ kJ mol$^{-1}$
Solution
$\Delta_{\text{sol}} H = \Delta_{\text{lattice}} H + \Delta_{\text{hyd}} H$; $4 = 788 + \Delta_{\text{hyd}} H$; $\Delta_{\text{hyd}} H = -784$ kJ/mol.
BETA
