Math Problem Statement

Hydrogen chloride and oxygen react to form chlorine and water: 4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g). A chemist finds that at a certain temperature the equilibrium mixture has the following pressures: HCl = 72.5 atm, O2 = 57.3 atm, Cl2 = 34.1 atm, H2O = 61.6 atm. Calculate the equilibrium constant Kp.

Solution

The problem asks you to calculate the equilibrium constant KpK_p for the given reaction at equilibrium:

4HCl(g)+O2(g)2Cl2(g)+2H2O(g)4 \, \text{HCl}(g) + \text{O}_2(g) \rightleftharpoons 2 \, \text{Cl}_2(g) + 2 \, \text{H}_2\text{O}(g)

The pressures of the gases at equilibrium are provided:

  • HCl\text{HCl}: 72.5 atm
  • O2\text{O}_2: 57.3 atm
  • Cl2\text{Cl}_2: 34.1 atm
  • H2O\text{H}_2\text{O}: 61.6 atm

The equilibrium constant KpK_p is calculated using the expression:

Kp=(PCl2)2(PH2O)2(PHCl)4(PO2)K_p = \frac{(P_{\text{Cl}_2})^2 (P_{\text{H}_2\text{O}})^2}{(P_{\text{HCl}})^4 (P_{\text{O}_2})}

Where:

  • PCl2P_{\text{Cl}_2} is the equilibrium pressure of chlorine
  • PH2OP_{\text{H}_2\text{O}} is the equilibrium pressure of water vapor
  • PHClP_{\text{HCl}} is the equilibrium pressure of hydrogen chloride
  • PO2P_{\text{O}_2} is the equilibrium pressure of oxygen

Substitute the given pressures:

Kp=(34.1)2×(61.6)2(72.5)4×(57.3)K_p = \frac{(34.1)^2 \times (61.6)^2}{(72.5)^4 \times (57.3)}

Let's calculate:

  1. (34.1)2=1162.81(34.1)^2 = 1162.81
  2. (61.6)2=3794.56(61.6)^2 = 3794.56
  3. (72.5)4=27631406.25(72.5)^4 = 27631406.25
  4. Multiply the numerator: 1162.81×3794.56=4412280.611162.81 \times 3794.56 = 4412280.61
  5. Multiply the denominator: 27631406.25×57.3=1583533554.62527631406.25 \times 57.3 = 1583533554.625
  6. Now divide:

Kp=4412280.611583533554.6250.002785K_p = \frac{4412280.61}{1583533554.625} \approx 0.002785

Rounded to two significant digits:

Kp0.0028K_p \approx 0.0028

Let me know if you'd like any clarifications or further details!


Here are 5 related questions:

  1. How does the equilibrium constant KpK_p change with temperature?
  2. What is the difference between KcK_c and KpK_p?
  3. How would adding more O2\text{O}_2 affect the equilibrium of this reaction?
  4. Why are pressure values used instead of concentrations in this KpK_p calculation?
  5. How does Le Chatelier's principle apply to this reaction?

Tip: Always double-check units when calculating equilibrium constants, as incorrect units can lead to wrong results.

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Math Problem Analysis

Mathematical Concepts

Equilibrium constants
Partial pressures

Formulas

Kp = (P_Cl2)^2 * (P_H2O)^2 / [(P_HCl)^4 * (P_O2)]

Theorems

Equilibrium law for gas-phase reactions

Suitable Grade Level

Undergraduate (College-level Chemistry)