Calculate the partial pressure (in atm) of HI at equilibrium when 1.64 atm of CH₃I and 1.64 atm of H₂O react at 3000 K according to the following chemical equatio CH₃I(g)+H₂O(g)⇌CH₃OH(g)+HI(g)K p=3.07×10 Report your answer to three significant figures in scientific notation.

Question

24-02-2024
Chemistry

High School

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Calculate the partial pressure (in atm) of HI at equilibrium when 1.64 atm of CH₃I and 1.64 atm of H₂O react at 3000 K according to the following chemical equatio CH₃I(g)+H₂O(g)⇌CH₃OH(g)+HI(g)K p=3.07×10 Report your answer to three significant figures in scientific notation.

Answer :

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halakukhan1315 halakukhan1315 · Answer 1 · 2025-02-21T11:38:47-05:00

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.

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