DeltaH^(0) for a reaction F_(2) + 2HCl rarr 2HF + Cl_(2) is given as -352.8 kJ. Delta H_(f)^(0) for HF is -268.3 KJ mol^(-1), then Delta H_(f)^(0) of HCl would be

DeltaH^(0) for a reaction F_(2) + 2HCl rarr 2HF + Cl_(2) is given as -

1.	DeltaH^(0) for a reaction F_(2) + 2HCl rarr 2HF + Cl_(2) is given as -352.8 kJ. Delta H_(f)^(0) for HF is -268.3 KJ mol^(-1), then Delta H_(f)^(0) of HCl would be
Answer» SOLUTION :`Delta H = H _(P) - H _(R) = 2 H _(HF) - 2H _(HCl)` `=- 352.8 kJ = 2 (0268.3) - 2 (H _(HCl))` `2 (H _(HCl))=-536.6 + 352.8 =-183.8kJ.` (or) `H _(HCl) =-91.9 kJ mol ^(-1).`