Consider the following equilibrium at 388.6 K: $N H_{4} H S (s) ⇌ N H_{3} (g) + H_{2} S (g)$ $NH_4HS(s) rightleftharpoons NH_3(g) + H_2S(g)$ . The partial pressure of each gas is 0.203 atm. How do you calculate $K_{P}$ $K_P$ and $K_{C}$ $K_C$ for the reaction?

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Consider the following equilibrium at 388.6 K: $N H_{4} H S (s) ⇌ N H_{3} (g) + H_{2} S (g)$ $NH_4HS(s) rightleftharpoons NH_3(g) + H_2S(g)$ . The partial pressure of each gas is 0.203 atm. How do you calculate $K_{P}$ $K_P$ and $K_{C}$ $K_C$ for the reaction?

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Answer 1 · 2017-03-13T09:08:59

$K_{P} = 0.203$ $K_P = 0.203$
( $atm$ $"atm"$ )

$K_{C} = 6.37 \times 10^{- 3}$ $K_C = 6.37 xx 10^(-3)$
( $M$ $"M"$ )

The first thing you can do is write out $K_{P}$ $K_P$ .

$K_{P} = \frac{P_{N H_{3}} P_{H_{2} S}}{P_{N H_{4} H S}}$ $K_P = (P_(NH_3)P_(H_2S))/(P_(NH_4HS))$

where $P_{i}$ $P_i$ is the partial pressure of gas $i$ $i$ .

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

$K_{P} = \frac{(0.203 atm) (0.203 atm)}{(0.203 atm)}$ $color(blue)(K_P) = (("0.203 atm")("0.203 atm"))/(("0.203 atm"))$

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

Converting to $K_{C}$ $K_C$ , we assume all the gases dealt with are ideal gases, so that we can use the ideal gas law:

$P V = n R T$ $PV = nRT$

Given that $\frac{n_{i}}{V_{i}} = [i]$ $n_i/V_i = [i]$ for a given gas $i$ $i$ , we can substitute all the pressure terms in $K_{P}$ $K_P$ with concentration terms as follows. For a given gas reaction

$a A + b B \to c C + d D$ $aA + bB -> cC + dD$ ,

we would have written

$K_{P} = \frac{P_{C}^{c} P_{D}^{d}}{P_{A}^{a} P_{B}^{b}}$ $K_P = (P_C^cP_D^d)/(P_A^aP_B^b)$ .

Thus, by substituting

$P_{i}^{n_{i}} = {(\frac{n_{i} R T}{V_{i}})}_{i}^{n_{i}} = {([i] R T)}_{i}^{n_{i}}$ $P_i^(n_i) = ((n_iRT)/V_i)^(n_i) = ([i]RT)^(n_i)$ ,

we then have:

$K_{P} = \frac{{([C] R T)}^{c} {([D] R T)}^{d}}{{([A] R T)}^{a} {([B] R T)}^{b}}$ $K_P = (([C]RT)^(c)([D]RT)^(d))/(([A]RT)^(a)([B]RT)^(b))$

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

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.

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Consider the following equilibrium at 388.6 K: $N H_{4} H S (s) ⇌ N H_{3} (g) + H_{2} S (g)$ $NH_4HS(s) rightleftharpoons NH_3(g) + H_2S(g)$ . The partial pressure of each gas is 0.203 atm. How do you calculate $K_{P}$ $K_P$ and $K_{C}$ $K_C$ for the reaction?

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Consider the following equilibrium at 388.6 K: $N H_{4} H S (s) ⇌ N H_{3} (g) + H_{2} S (g)$ $NH_4HS(s) rightleftharpoons NH_3(g) + H_2S(g)$ . The partial pressure of each gas is 0.203 atm. How do you calculate $K_{P}$ $K_P$ and $K_{C}$ $K_C$ for the reaction?