Question

Consider the following equilibrium, for which Kc = 515 at 25.3 °C:

N2(g)+O2(g)+Br2(g)→2NOBr(g)

What is the value of Kp for this reaction?

Note: Use 3 signficant figures in your response.

Answer 1

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

\[
\boxed{21.0}
\]