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At a certain temperature, the Kp for the decomposition of H₂S is 0.700. H₂S(g) H₂(g) + S(g) Initially, only H₂S is present at a pressure of 0.217 atm in a closed container. What is the total pressure in the container at equilibrium? =

At a certain temperature, the Kp for the decomposition of H₂S is 0.700. H₂S(g) H₂(g) + S(g) Initially, only H₂S is present at a pressure of 0.217 atm in a closed container. What is the total pressure in the container at equilibrium? =

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Balanced Reaction for the decomposition of H₂Sis as follows H₂S (g) — H₂(g) + S (g) Equilibrium constant (K₂) =0.800 Initial pressure of H₂S= 0.271 atm Set an ICE table we can write H₂S (g) → H₂(g) + S (g) I 0.271 0 C -X +X E (0.271-x) X Where, x= change in partial pressure. So we can write the Kp as follows PH₂Ps PH₂S Kp = +X X Putting values we can write x² 0.800 XXX 0.271-x 0.271-x x² +0.800x - 0.2168 = 0 Now, calculate the value of x from this equation we can write = V X = -0.800+√0.800²+4×1×0.2168 2×1 X = = 0.214and x = -1.014 Now, x can not be negative so, x will be 0.214 Explanation Please refer to solution in this step. Equilibrium partial pressure of H₂S = (0.271-0.214)=0.057 atm Partial pressure of H₂=0.214 atm Partial pressure of S=0.214 atm Total partial pressure Ptotal= (0.057+0.214+0.214)atm =0.485 atm
Transcribed Image Text:Balanced Reaction for the decomposition of H₂Sis as follows H₂S (g) — H₂(g) + S (g) Equilibrium constant (K₂) =0.800 Initial pressure of H₂S= 0.271 atm Set an ICE table we can write H₂S (g) → H₂(g) + S (g) I 0.271 0 C -X +X E (0.271-x) X Where, x= change in partial pressure. So we can write the Kp as follows PH₂Ps PH₂S Kp = +X X Putting values we can write x² 0.800 XXX 0.271-x 0.271-x x² +0.800x - 0.2168 = 0 Now, calculate the value of x from this equation we can write = V X = -0.800+√0.800²+4×1×0.2168 2×1 X = = 0.214and x = -1.014 Now, x can not be negative so, x will be 0.214 Explanation Please refer to solution in this step. Equilibrium partial pressure of H₂S = (0.271-0.214)=0.057 atm Partial pressure of H₂=0.214 atm Partial pressure of S=0.214 atm Total partial pressure Ptotal= (0.057+0.214+0.214)atm =0.485 atm
At a certain temperature, the Kp for the decomposition of H₂S is 0.700. H₂S(g) = H₂(g) + S(g) Initially, only H₂S is present at a pressure of 0.217 atm in a closed container. What is the total pressure in the container at equilibrium?
Transcribed Image Text:At a certain temperature, the Kp for the decomposition of H₂S is 0.700. H₂S(g) = H₂(g) + S(g) Initially, only H₂S is present at a pressure of 0.217 atm in a closed container. What is the total pressure in the container at equilibrium?
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