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Ch.15 - Chemical Equilibrium
Chapter 15, Problem 82

Phosphine (PH3) decomposes at elevated temperatures, yielding gaseous P2 and H2: 2 PH3(g) ⇌ P2(g) + 3 H2(g), Kp = 398 at 873 K. (b) When a mixture of PH3, P2, and H2 comes to equilibrium at 873 K, P_P2 = 0.412 atm and P_H2 = 0.822 atm. What is P_PH3?

Verified step by step guidance
1
Step 1: Write the expression for the equilibrium constant Kp for the reaction: Kp = (P_{P2} * (P_{H2})^3) / (P_{PH3})^2.
Step 2: Substitute the given values into the Kp expression: Kp = 398, P_{P2} = 0.412 atm, and P_{H2} = 0.822 atm.
Step 3: Rearrange the Kp expression to solve for P_{PH3}: (P_{PH3})^2 = (P_{P2} * (P_{H2})^3) / Kp.
Step 4: Substitute the known values into the rearranged equation: (P_{PH3})^2 = (0.412 * (0.822)^3) / 398.
Step 5: Solve for P_{PH3} by taking the square root of the result from Step 4.
Related Practice
Open Question
For each of the following equilibria, write the equilibrium constant expression for Kc. Where appropriate, also write the equilibrium constant expression for Kp. (a) WO3(s) + 3 H2(g) ⇌ W(s) + 3 H2O(g) (b) Ag+(aq) + Cl-(aq) ⇌ AgCl(s) (c) 2 FeCl3(s) + 3 H2O(g) ⇌ Fe2O3(s) + 6 HCl(g) (d) MgCO3(s) ⇌ MgO(s) + CO2(g)
Textbook Question
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Textbook Question
The value of Kc for the reaction 3 O21g2 ∆ 2 O31g2 is 1.7 * 10-56 at 25°C. Do you expect pure air at 25 °C to contain much O3 (ozone) when O2 and O3 are in equilib- rium? If the equilibrium concentration of O2 in air at 25 °C is 8 * 10-3 M, what is the equilibrium concentration of O3?
601
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Open Question
At 500 K, the equilibrium constant for the dissociation reaction H2(g) ⇌ 2H(g) is very small (Kc = 1.2 × 10⁻⁴²). (a) What is the molar concentration of H atoms at equilibrium if the equilibrium concentration of H2 is 0.10 M? (b) How many H atoms and H2 molecules are present in 1.0 L of 0.10 M H2 at 500 K?
Open Question
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Textbook Question
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