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"#)
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.
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Answer 2

To calculate ( K_P ):

[ K_P = \frac{{P_{NH_3} \cdot P_{H_2S}}}{{P_{NH_4HS}}} ]

To calculate ( K_C ):

[ K_C = \frac{{[NH_3] \cdot [H_2S]}}{{[NH_4HS]}} ]

Given that the partial pressure of each gas is 0.203 atm, use these values to calculate ( K_P ) and ( K_C ).

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Answer 3

To calculate ( K_P ) and ( K_C ) for the reaction ( NH_4HS(s) \rightleftharpoons NH_3(g) + H_2S(g) ) at 388.6 K, we first need to understand the equilibrium expression for both forms.

For ( K_P ), the equilibrium constant in terms of partial pressures is given by:

[ K_P = \frac{{(P_{NH_3} \times P_{H_2S})}}{{P_{NH_4HS}}} ]

Given that the partial pressure of each gas is 0.203 atm, we substitute these values into the equation to find ( K_P ):

[ K_P = \frac{{(0.203 \times 0.203)}}{{P_{NH_4HS}}} ]

For ( K_C ), the equilibrium constant in terms of concentrations is given by:

[ K_C = \frac{{[NH_3] \times [H_2S]}}{{[NH_4HS]}} ]

Since ( NH_4HS ) is a solid, its concentration remains constant and can be omitted from the expression. Therefore, ( K_C ) simplifies to:

[ K_C = \frac{{[NH_3] \times [H_2S]}}{{[NH_4HS]}} ]

Given that the partial pressure of each gas is 0.203 atm, and under ideal gas conditions, pressure is directly proportional to concentration, we can substitute the partial pressures directly into the equation:

[ K_C = \frac{{(0.203 \times 0.203)}}{{[NH_4HS]}} ]

It's important to note that the concentration of ( NH_4HS ) doesn't change because it's a solid, so it effectively stays constant.

Therefore, to calculate both ( K_P ) and ( K_C ), you would use the given partial pressures of 0.203 atm for ( NH_3 ) and ( H_2S ), and then perform the necessary calculations as outlined above.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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