ammonium hydrogen sulfide decomposes according to the reaction.
NH4HS(s) ⇌ NH3(g) + H2S(g) Kc=0.11 at 250°C
If 55.0g of solid NH4HS is placed in a sealed 5.0L container,what is the partial pressure of NH3 and H2S at equilibrium?
NH4HS(s) ⇌ NH3(g) + H2S(g) Kc=0.11 at 250°C
If 55.0g of solid NH4HS is placed in a sealed 5.0L container,what is the partial pressure of NH3 and H2S at equilibrium?
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First convert Kc to Kp.
Kp=Kc(RT)∆n
Kp=0.11(0.0821*523)^-1=2.56*10^-3
Find whether the reaction will shift to products or reactants.
Use coefficient of product for numerator and coefficient of reactant for the denominator.
Q=1*1/1
Q=1
Since 1>2.56*10^-3, the reaction will shift to create more products. This determines the +/- of the unknown amounts. Initial atm's are not given for the products so just set Kp (2.56*10^-3) equal to the change in products over the change in reactants. Solids are not included since their concentration never changes (neither do pure liquids) so we omit it in the equation.
2.56*10^-3=(x*x)
x=5.06x10^-2 atm's for each product at equilibrium.
Kp=Kc(RT)∆n
Kp=0.11(0.0821*523)^-1=2.56*10^-3
Find whether the reaction will shift to products or reactants.
Use coefficient of product for numerator and coefficient of reactant for the denominator.
Q=1*1/1
Q=1
Since 1>2.56*10^-3, the reaction will shift to create more products. This determines the +/- of the unknown amounts. Initial atm's are not given for the products so just set Kp (2.56*10^-3) equal to the change in products over the change in reactants. Solids are not included since their concentration never changes (neither do pure liquids) so we omit it in the equation.
2.56*10^-3=(x*x)
x=5.06x10^-2 atm's for each product at equilibrium.