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.

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

\[
K_p = K_c (RT)^{\Delta n}
\]

Where:
- \(K_p\) is the equilibrium constant in terms of pressure.
- \(K_c\) is the equilibrium constant in terms of concentration.
- \(R\) is the gas constant (\(0.0821 \, \text{L·atm/mol·K}\)).
- \(T\) is the temperature in Kelvin (for 25°C, \(T = 298 \, \text{K}\)).
- \(\Delta n\) is the change in the number of moles of gas, calculated as:

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

### Step 1: Calculate \(\Delta n\)
The balanced reaction is:
\[
4 \, \text{NO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \, \text{N}_2\text{O}_5(g)
\]
- Moles of gaseous products = 2 (from \(2 \, \text{N}_2\text{O}_5\))
- Moles of gaseous reactants = 4 + 1 = 5 (from \(4 \, \text{NO}_2\) and \(1 \, \text{O}_2\))

So, \(\Delta n = 2 - 5 = -3\).

### Step 2: Rearrange the formula to solve for \(K_c\)
\[
K_c = \frac{K_p}{(RT)^{\Delta n}}
\]

### Step 3: Plug in the values
- \(K_p = 7.66 \times 10^{-6}\)
- \(R = 0.0821 \, \text{L·atm/mol·K}\)
- \(T = 298 \, \text{K}\)
- \(\Delta n = -3\)

Now, calculate:

\[
K_c = \frac{7.66 \times 10^{-6}}{(0.0821 \times 298)^{-3}}
\]

Let me calculate this for you.The value of \(K_c\) 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 \(K_p\) and \(K_c\)?
2. What happens to the equilibrium position if the pressure of this reaction is increased?
3. Can you explain why \(\Delta n\) affects the relationship between \(K_p\) and \(K_c\)?
4. What are the typical units of \(K_p\) and \(K_c\), and how do they differ?
5. How would you calculate \(K_p\) if \(K_c\) were given instead?

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

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

Learn how to convert equilibrium constants Kp to Kc for the reaction 4 NO2(g) + O2(g) ⇌ 2 N2O5(g) at 25°C. This problem involves using the formula Kp = Kc(RT)^Δn, where Δn represents the change in moles of gas, and provides a step-by-step solution. Suitable for advanced chemistry students.

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