`HCI +NaOH rarr naCI +H_(2)O, DeltaH =- 57.1 kJ mol^(-1)`
`DeltaH` is the heat of ……………….of `HCI` and `NaOH` solution.

H_(2)O(l) rarr H_(2)O(s), DeltaH =- 6.01 kJ DeltaH is the heat of ……………….of water.

H_(2)O(s) rarr H_(2)O(l), DeltaH = 6.01 kJ DeltaH is the heat of …………………of ice.

H_(2)O(g) rarr H_(2)O(l), DeltaH =- 40.7 kJ DeltaH is the heat of……………….of water.

H_(2)O(l) rarr H_(2)O(g), DeltaH = +40.7 kJ DeltaH is the heat of ……………..of water.

CsOH +HCI rarr CsCI +H_(2)O, DeltaH =- 13.4 kcal mol^(-1)……….(i) CsOH +HF rarr CsF +H_(2)O, DeltaH =- 16.4 kcal mol^(-1) ………(ii) Calculate DeltaH for the ionisation of HF in H_(2)O .

C_(4)H_(10)+(13)/(2)O_(2)(g)rarr4CO_(2)(g)+5H_(2)O(l), DeltaH =- 2878 kJ DeltaH is the heat of……….of butane gas.

Given CaCI_(2)(s) +aq rarr CaCI_(2)(aq): DeltaH^(Theta) = 75 kJ mol^(-1) at 18^(@)C and CaCI_(2).6H_(2)O +aq rarr CaCI_(2)(aq), DeltaH^(Theta) 19 kJ mol^(-1) at 18^(@)C . Find the heat of hydration of CaCI_(2) to CaCI_(2). 6H_(2)O by H_(2)O (g) . The Heat of vaporisation of water may be taken as 2452 J^(-1)g at 18^(@)C .

Heat of neutralisation for the reaction: NaOH+HCl to NaCl+H_(2)O is 57.1 kJ mol^(-1) . The heat released when 0.25 mole of NaOH is titrated against 0.25 mole of HCl will be:

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)