Calculate `K_(sp)` for `AgCI`. Given:
`Delta_(f)H^(Θ) Ag^(o+) = 25.3 kcal mol^(-1)`
`Delta_(f)H^(Θ) C1^(Θ) =- 40.0 kcal mol^(-1)`
`Delta_(f)H^(Θ) AgC1=- 30.36 kcal mol^(-1)`
`S^(Θ) Ag^(o+), S^(Θ) C1^(Θ)`, and `S^(Θ) AgC1`are `17.7, 13.2` and `23.0 cal mol^(-1)`

Calculate Deltah for the eaction BaCO_(3)(s)+2HCI(aq) rarr BaCI_(2)(aq)+CO_(2)(g)+H_(2)O(l) Delta_(f)H^(Theta) (BaCO_(3)) =- 290.8 kcal mol^(-1), Delta_(f)H^(Theta) (H^(o+)) =0 Delta_(f)H^(Theta) (Ba^(++)) = - 128.67 kcal mol^(-1) , Delta_(f)H^(Theta)(CO_(2)) =- 94.05 kcal mol^(-1) , Delta_(f)H^(Theta) (H_(2)O) =- 68.32 kcal mol^(-1)

Calculate the enthalpy change for the process C Cl_(4)(g) rarr C(g)+4Cl(g) and calculate bond enthalpy of C-Cl in C Cl_(4)(g) . Delta_(vap)H^(Θ)(C Cl_(4))=30.5 kJ mol^(-1) Delta_(f)H^(Θ)(C Cl_(4))=-135.5 kJ mol^(-1) Delta_(a)H^(Θ)(C )=715.0 kJ mol^(-1) , where Delta_(a)H^(Θ) is enthalpy of atomisation Delta_(a)H^(Θ)(Cl_(2))=242 kJ mol^(-1)

Calculate the lattice energy from the following data (given 1 eV = 23.0 kcal mol^(-1) ) i. Delta_(f) H^(ɵ) (KI) = -78.0 kcal mol^(-1) ii. IE_(1) of K = 4.0 eV iii. Delta_("diss")H^(ɵ)(I_(2)) = 28.0 kcal mol^(-1) iv. Delta_("sub")H^(ɵ)(K) = 20.0 kcal mol^(-1) ltbvrgt v. EA of I = -70.0 kcal mol^(-1) vi. Delta_("sub")H^(ɵ) of I_(2) = 14.0 kcal mol^(-1)

From the following data, calculate the standard enthalpy of formation of propane Delta_(f)H^(Theta) CH_(4) = - 17 kcal mol^(-1) Delta_(f)H^(Theta)C_(2)H_(6) =- 24 kcal mol^(-1), BE (C-H) = 99 kcal mol^(-1) (C- C) = 84 kcal mol^(-1) .

Estimate the K_(sp) of AgBr . Given Delta_(f)H^(Θ) of Ag^(o+), Br^(Θ) , and AgBr are 25.31,-28.9 , and -23.8kcal, DeltaS^(Θ) of Ag^(o+), Br^(Θ) , and AgBr are 17.7, 19.3, and -26.6 cal//K .

Calculate the electronegativity of fluorine from the following data: E_(H-H) = 104.4 kcal mol^(-1), E_(F-F) = 36.6 kcal mol^(-1) E_(H-F) = 134.3 kcal mol^(-1), chi_(H) = 2.1

Calculate the electronegativity of fluorine from following data : E_(H - H) = 104.2 kcal mol^(-1) E_(F - F) = 36.6 kcal mol^(-1) E_(H -F) = 134.6 kcal mol^(-1) Electronegativity of H is 2.05.

Delta_(f) H of graphite is 0.23 kJ mol^(-1) and Delta_(f) H of diamond is 1.896 kh mol^(-1) . Delta H_(transition) from graphite to diamond is

Find DeltaH of the process NaOH(s) rarr NaOH(g) Given: Delta_(diss)H^(Theta)of O_(2) = 151 kJ mol^(-1) Delta_(diss)H^(Theta) of H_(2) = 435 kJ mol^(-1) Delta_(diss)H^(Theta) of O-H = 465 kJ mol^(-1) Delta_(diss)H^(Theta) of Na -O = 255 kJ mol^(-1) Delta_(soln)H^(Theta)of NaOH = - 46 kJ mol^(-1) Delta_(f)H^(Theta) of NaOH(s) =- 427 kJ mol^(-1) Delta_("sub")H^(Theta) of Na(s) = 109 kJ mol^(-1)