At `298K`, if `/_\G_(f)^(@)` of `HCL_((g))` is `1.72kJmol^(-1)`, then calculate `K_(p)` for the following reversible reaction: `2HCl_((g))hArrH_(2(g))+Cl_(2(g))`
(use: at `298K:2.303RT=5700Jmol^(-1)` and `log2=0.30`)

Answer

4

Explanation

`[4]`
`1/2H_(2)(g)+1/2Cl_(2)(g) hArr(HCl)(g)/_\G^(@)=1.72kJmol^(-1)`
`/_\G^(@)` for `2HCl (g) hArrH_(2)(g)+Cl_(2)(g)` is `[2xx(-1.72)]kJ`
`/_\G^(@)=-2.303RTlogK_(p)`
`-3.44xx10^(3)=-5700logK_(p)`
`logK_(p)=0.6`
`K_(p)=4`