Question

Consider the following equilibrium at 388.6 K: `NH_4HS(s) rightleftharpoons NH_3(g) + H_2S(g)`. The partial pressure of each gas is 0.203 atm. How do you calculate `K_P` and `K_C` for the reaction?

Answer 1

`K_P = 0.203`
(`"atm"`)

`K_C = 6.37 xx 10^(-3)`
(`"M"`)

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The first thing you can do is write out `K_P`.

`K_P = (P_(NH_3)P_(H_2S))/(P_(NH_4HS))`

where `P_i` is the partial pressure of gas `i`.

The stoichiometries are all 1:1:1, so all the exponents are `1`. Thus, as all partial pressures are given, we have a very simple equilibrium expression to evaluate:

`color(blue)(K_P) = (("0.203 atm")("0.203 atm"))/(("0.203 atm"))`

`=` `color(blue)("0.203")`
`" "`(`"atm"`)

Converting to `K_C`, we assume all the gases dealt with are ideal gases, so that we can use the ideal gas law:

`PV = nRT`

Given that `n_i/V_i = [i]` for a given gas `i`, we can substitute all the pressure terms in `K_P` with concentration terms as follows. For a given gas reaction

`aA + bB -> cC + dD`,

we would have written

`K_P = (P_C^cP_D^d)/(P_A^aP_B^b)`.

Thus, by substituting

`P_i^(n_i) = ((n_iRT)/V_i)^(n_i) = ([i]RT)^(n_i)`,

we then have:

`K_P = (([C]RT)^(c)([D]RT)^(d))/(([A]RT)^(a)([B]RT)^(b))`

`= stackrel(K_C)overbrace(([C]^c[D]^d)/([A]^a[B]^b))((RT)^(c)(RT)^(d))/((RT)^a(RT)^b)`

At this point we've separated out the very definition of `K_C`. Now, just use the properties of exponents to condense the `RT` terms together.

`= K_C((RT)^(c+d))/((RT)^(a+b))`

`= K_C(RT)^((c+d)-(a+b))`

But since the stoichiometries of a reaction can be related through the moles of reactants and products, we might as well call the exponent

`Deltan_(gas) = (n_c + n_d) - (n_a + n_b)`,

where `Deltan_(gas)` is the mols of product gases minus the mols of reactant gases.

Therefore, to convert from `K_P` to `K_C`, we simply have:

`color(green)(K_P = K_C(RT)^(Deltan_(gas)))`

For this expression, be sure to use `bb(R = "0.08206 L"cdot"atm/mol"cdot"K")`. To get `K_C` then, we simply have:

`color(blue)(K_C) = (K_P)/((RT)^(Deltan_(gas)))`

`= (0.203 cancel"atm")/[("0.08206 L"cdotcancel"atm""/mol"cdotcancel"K")(388.6 cancel"K")]^((1 + 1) - (1))`

`= color(blue)(6.37 xx 10^(-3))`

Technically, the units are `("mol"/"L")^((1+1) - (1)) = "mol"/"L"`, but generally `K_C` is reported without units.