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?

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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?

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Md Ariful Islam · Answer 1 · Apr 02, 2024 09:48:30

$K_P = 0.203$
($"atm"$)

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

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 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 :

$PV = nRT$

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.

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?

1 Answers