# 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?

we would have written

Thus, by substituting

we then have:

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

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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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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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