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For the reaction : F(2)(g) + 2HCl(g) h...

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 :

A

`- 0.603 kcalg^(-1)`

B

`0.603 kcalg^(-1)`

C

`0.0603 kcalg^(-1)`

D

`6.03 kcalg^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A

`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`
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