Math Problem Statement

4 NO2(g) + O2(g) equilibrium reaction arrow 2 N2O5(g) Kp = 7.66 ✕ 10−6 at 25°CConvert the values of Kc to values of Kp or the values of Kp to values of Kc.

Solution

To convert between the equilibrium constants KpK_p (based on partial pressures) and KcK_c (based on concentrations) for a gas-phase reaction, we use the following formula:

Kp=Kc(RT)ΔnK_p = K_c (RT)^{\Delta n}

Where:

  • KpK_p is the equilibrium constant in terms of pressure.
  • KcK_c is the equilibrium constant in terms of concentration.
  • RR is the gas constant (0.0821L\cdotpatm/mol\cdotpK0.0821 \, \text{L·atm/mol·K}).
  • TT is the temperature in Kelvin (for 25°C, T=298KT = 298 \, \text{K}).
  • Δn\Delta n is the change in the number of moles of gas, calculated as:

Δn=(moles of gaseous products)(moles of gaseous reactants)\Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})

Step 1: Calculate Δn\Delta n

The balanced reaction is: 4NO2(g)+O2(g)2N2O5(g)4 \, \text{NO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \, \text{N}_2\text{O}_5(g)

  • Moles of gaseous products = 2 (from 2N2O52 \, \text{N}_2\text{O}_5)
  • Moles of gaseous reactants = 4 + 1 = 5 (from 4NO24 \, \text{NO}_2 and 1O21 \, \text{O}_2)

So, Δn=25=3\Delta n = 2 - 5 = -3.

Step 2: Rearrange the formula to solve for KcK_c

Kc=Kp(RT)ΔnK_c = \frac{K_p}{(RT)^{\Delta n}}

Step 3: Plug in the values

  • Kp=7.66×106K_p = 7.66 \times 10^{-6}
  • R=0.0821L\cdotpatm/mol\cdotpKR = 0.0821 \, \text{L·atm/mol·K}
  • T=298KT = 298 \, \text{K}
  • Δn=3\Delta n = -3

Now, calculate:

Kc=7.66×106(0.0821×298)3K_c = \frac{7.66 \times 10^{-6}}{(0.0821 \times 298)^{-3}}

Let me calculate this for you.The value of KcK_c is approximately 0.112 at 25°C.

Would you like further details or have any questions?

Here are 5 related questions to deepen understanding:

  1. How does the change in temperature affect the relationship between KpK_p and KcK_c?
  2. What happens to the equilibrium position if the pressure of this reaction is increased?
  3. Can you explain why Δn\Delta n affects the relationship between KpK_p and KcK_c?
  4. What are the typical units of KpK_p and KcK_c, and how do they differ?
  5. How would you calculate KpK_p if KcK_c were given instead?

Tip: Always remember that temperature must be in Kelvin when using gas laws or equilibrium constants.

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Math Problem Analysis

Mathematical Concepts

Equilibrium Constants
Gas Laws
Chemical Kinetics

Formulas

Kp = Kc (RT)^Δn
Kc = Kp / (RT)^Δn

Theorems

Ideal Gas Law

Suitable Grade Level

Grades 11-12 (Advanced Chemistry)