Questions: Consider the reaction below: 2 CO(g) + O2(g) ⇌ 2 CO2(g) If Kc is 2.24 × 10^22 at 1273.0°C, calculate Kp at the same temperature. (R= 0.08206 L · atm / mol · K.)

Solution Steps

The relationship between the equilibrium constants \( K_c \) and \( K_p \) for a gaseous reaction is given by the equation:

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

where:

• \( R \) is the ideal gas constant, \( 0.08206 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \),
• \( T \) is the temperature in Kelvin,
• \( \Delta n \) is the change in moles of gas, calculated as the moles of gaseous products minus the moles of gaseous reactants.

For the given reaction:

\[ 2 \mathrm{CO}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}_{2}(\mathrm{~g}) \]

• Moles of gaseous products = 2 (from \( \mathrm{CO}_2 \))
• Moles of gaseous reactants = 2 (from \( \mathrm{CO} \)) + 1 (from \( \mathrm{O}_2 \)) = 3

Thus,

\[ \Delta n = 2 - 3 = -1 \]

The temperature given is \( 1273.0^{\circ} \mathrm{C} \). Convert this to Kelvin:

\[ T = 1273.0 + 273.15 = 1546.15 \, \text{K} \]

Substitute the values into the equation for \( K_p \):

\[ K_p = K_c \left( RT \right)^{\Delta n} = 2.24 \times 10^{22} \left( 0.08206 \times 1546.15 \right)^{-1} \]

Calculate \( RT \):

\[ RT = 0.08206 \times 1546.15 = 126.8511 \]

Now calculate \( K_p \):

\[ K_p = 2.24 \times 10^{22} \times (126.8511)^{-1} \]

\[ K_p = 2.24 \times 10^{22} \times 0.007882 \]

\[ K_p = 1.764 \times 10^{20} \]

Final Answer