Question

I calculated the partial pressure to be 16.1 - and it was wrong?

Consider the following equilibrium reaction at `256^circC`.
`SbCl_(5"(g)")⇌SbCl_("3(g)") + Cl_(2"(g)")`

If the equilibrium constant in terms of concentrations (`K_c`) is `3.7 xx 10^-2`, calculate the equilibrium constant in terms of partial pressures (`K_p`).

Answer 1

`K_p~~163`

Explanation:

`K_c` and `K_p` are linked by the following reaction:
`K_p=K_c(RT)^(Deltan)`, where:

• `K_c` = equilibrium constant in terms of pressure
• `R` = gas constant (`8.31"J"` `"K"^-1` `"mol"^-1`)
• `T` = absolute temperature (`K`)
• `Deltan="Number of moles of gaseous products"-"Number of moles of gaseous reactants"`

`K_c=3.7*10^-2`
`R=8.31"J"` `"K"^-1` `"mol"^-1`
`T=256+273=529K`
`Deltan=1`

`K_p=(3.7*10^-2)(529*8.31)^1=(3.7*10^-2)(529*8.31)=162.65163~~163`