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NEET Chemistry QUESTION #1083
Question 1
The standard heat of formation of Ba\(^{2+}\)(aq) in kcal/mol is: [Given: \(\Delta H_f^\circ(\text{SO}_4^{2-}(\text{aq})) = -216\ \text{kcal/mol}\), \(\Delta H_\text{cryst}(\text{BaSO}_4) = -4.5\ \text{kcal/mol}\), \(\Delta H_f^\circ(\text{BaSO}_4(\text{s})) = -349\ \text{kcal/mol}\)]
  • \(+133.0\)
  • \(+220.5\)
  • \(-128.5\)
  • \(-133.0\)✔️
Correct Answer Explanation
\(\Delta H_f^\circ(\text{BaSO}_4,\text{s}) = \Delta H_f^\circ(\text{Ba}^{2+},\text{aq}) + \Delta H_f^\circ(\text{SO}_4^{2-},\text{aq}) + \Delta H_\text{cryst}\). \(-349 = \Delta H_f^\circ(\text{Ba}^{2+}) + (-216) + (-4.5)\). \(\Delta H_f^\circ(\text{Ba}^{2+}) = -349 + 216 + 4.5 = -128.5\ \text{kcal/mol}\). Wait: crystallisation is lattice formation from ions, so: \(\Delta H_f^\circ(\text{BaSO}_4,\text{s}) = \Delta H_f^\circ(\text{Ba}^{2+}) + \Delta H_f^\circ(\text{SO}_4^{2-}) + \Delta H_\text{cryst}\Rightarrow \Delta H_f^\circ(\text{Ba}^{2+})=-349+216+4.5=-128.5\). Hmm, but options show -133.0. Using Hess's: \(\Delta H_f^\circ(\text{Ba}^{2+})=-349-(-216)-(-4.5)=-349+216+4.5=-128.5\). Closest: option (3) = -128.5.