1.

For the reaction : F_(2)(g) + 2HCl(g) hArr 2HF(g) + Cl_(2)(g) Delta H^(theta) at 25^(@)C is =-84.4 kcalmol^(-1) Delta_(f) H^(theta)(HF) = -64.2kcalmol^(-1) Delta_(f) H^(theta) for HCl(g) per gram is :

Answer»

`- 0.603 kcalg^(-1)`
`0.603 kcalg^(-1)`
`0.0603 kcalg^(-1)`
`6.03 kcalg^(-1)`

Solution :`Delta H = 2 Delta H_(f) [HF(g)] + Delta H_(f)(Cl_(2)) - 2 Delta H_(f) [HCL(g)] - Delta H_(f)(F_(2))`
`- 84.4 = 2 xx (-64.2) - 0 - 2 Delta H_(f) [HCl(g)] - 0`
`2 Delta H_(f)[HCl(g)] = - 128.4 + 84.4 = - 44.0 KCAL`
`Delta H_(f) [HCl(g)] = - 22.0 kcal`
`Delta H_(f)^(@)` for HCl per GRAM `= (-22.0)/(36.5)`
`= - 0.603 kcal`


Discussion

No Comment Found

Related InterviewSolutions