Answer :
Final Answer:
The partial pressure of HI at equilibrium is 0.132786 atm 4.29E-2 atm.
Explanation:
At equilibrium, the reaction quotient (Qc) is equal to the equilibrium constant (Kp). The given equilibrium equation can be written as follows:
CH₃I(g) + H₂O(g) ⇌ CH₃OH(g) + HI(g)
At equilibrium, the number of moles of CH₃I and H₂O is equal to the number of moles of CH₃OH and HI.
According to the given values, the initial partial pressure of CH₃I and H₂O is 1.64 atm each. Therefore, the initial total pressure is 3.28 atm.
Using the equilibrium constant, Kp, which is given as 3.07 × 10, we can calculate the equilibrium partial pressures.
Kp = (PCH₃OH × PHI) / (PCH₃I × PH₂O)
Let's assume that the equilibrium partial pressure of HI is x atm.
Kp = [CH₃OH] [HI] / [CH₃I] [H₂O]
Kp = (x) / (1.64) (1.64)
Kp = x / 2.6896
Substituting the value of Kp, we get:
3.07 × 10 = x / 2.6896
x = 0.132786
Therefore, the partial pressure of HI at equilibrium is 0.132786 atm or 4.29E-2 atm, when reported to three significant figures in scientific notation.
Learn more about: equilibrium
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