Questions: Advanced General Chemistry Calculating an equilibrium constant from an equilibrium composition Morgan , hydrogen bromide and oxygen react to form bromine and water, like this: 4 HBr(g) + O2(g) -> 2 Br2(g) + 2 H2O(g) Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen bromide, oxygen, bromine, and water has the following composition: Compound Pressure at equilibrium ------ HBr 79.3 atm O2 60.9 atm Br2 80.9 atm H2O 21.7 atm Calculate the value of the equilibrium constant Kp for this reaction. Round your answer to 2 significant digits.

Solution Steps

For the given reaction:

\[ 4 \mathrm{HBr}(g) + \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{Br}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(g) \]

The equilibrium constant \( K_p \) is given by the expression:

\[ K_p = \frac{(P_{\mathrm{Br}_2})^2 (P_{\mathrm{H}_2\mathrm{O}})^2}{(P_{\mathrm{HBr}})^4 (P_{\mathrm{O}_2})} \]

where \( P \) denotes the partial pressure of each gas at equilibrium.

Substitute the given equilibrium pressures into the expression for \( K_p \):

• \( P_{\mathrm{HBr}} = 79.3 \, \text{atm} \)
• \( P_{\mathrm{O}_2} = 60.9 \, \text{atm} \)
• \( P_{\mathrm{Br}_2} = 80.9 \, \text{atm} \)
• \( P_{\mathrm{H}_2\mathrm{O}} = 21.7 \, \text{atm} \)

\[ K_p = \frac{(80.9)^2 (21.7)^2}{(79.3)^4 (60.9)} \]

Calculate the value of \( K_p \) using the substituted values:

\[ K_p = \frac{(80.9)^2 \times (21.7)^2}{(79.3)^4 \times 60.9} \]

First, calculate the numerator:

\[ (80.9)^2 = 6544.81 \] \[ (21.7)^2 = 470.89 \] \[ 6544.81 \times 470.89 = 3080860.7609 \]

Next, calculate the denominator:

\[ (79.3)^4 = 39691296.01 \] \[ 39691296.01 \times 60.9 = 2416780074.209 \]

Finally, calculate \( K_p \):

\[ K_p = \frac{3080860.7609}{2416780074.209} \approx 0.001275 \]

Final Answer