To convert \( K_c \) to \( K_p \), we can use the following relationship:
\[ K_p = K_c (RT)^{\Delta n} \]
where:
- \( R \) is the universal gas constant, \( 0.0821 , \text{L atm/(K mol)} \)
- \( T \) is the temperature in Kelvin
- \( \Delta n \) is the change in the number of moles of gas between products and reactants.
First, we need to calculate \( \Delta n \):
For the given reaction:
\[ \text{N}_2(g) + \text{O}_2(g) + \text{Br}_2(g) \rightarrow 2 \text{NOBr}(g) \]
- Moles of gas on the reactant side = 1 (N2) + 1 (O2) + 1 (Br2) = 3
- Moles of gas on the product side = 2 (NOBr)
So,
\[ \Delta n = \text{moles of products} - \text{moles of reactants} = 2 - 3 = -1 \]
Now, we must convert the temperature \( 25.3 , ^\circ C \) to Kelvin:
\[ T = 25.3 + 273.15 = 298.45 , K \]
Now substituting the values into the equation, we have:
- \( K_c = 515 \)
- \( R = 0.0821 , \text{L atm/(K mol)} \)
- \( T = 298.45 , K \)
- \( \Delta n = -1 \)
Calculating \( K_p \):
\[ K_p = 515 \cdot (0.0821 , \text{L atm/(K mol)} \cdot 298.45 , K)^{-1} \]
Calculating \( RT \):
\[ RT = 0.0821 \times 298.45 \approx 24.478 , \text{L atm/mol} \]
Thus,
\[ K_p = 515 \cdot (24.478)^{-1} \]
Calculating \( K_p \):
\[ K_p \approx 515 \cdot \frac{1}{24.478} \approx 515 \cdot 0.0408 \approx 20.99 \]
So, rounded to three significant figures, the value of \( K_p \) is
\[ \boxed{21.0} \]