Certainly! Let's calculate the equilibrium constant [tex]\( K_p \)[/tex] for the given reaction at the given temperature.
First, let's write the balanced chemical equation:
[tex]\[ 2 \text{CO} (g) + \text{O}_2 (g) \rightleftharpoons 2 \text{CO}_2 (g) \][/tex]
Here, we are given [tex]\( K_c = 2.24 \times 10^{22} \)[/tex] at a temperature of [tex]\( 1273.0^\circ \text{C} \)[/tex].
To use the formula that relates [tex]\( K_c \)[/tex] and [tex]\( K_p \)[/tex], we'll need to convert the temperature from Celsius to Kelvin:
[tex]\[ T(K) = 1273.0^\circ \text{C} + 273.15 = 1546.15\ \text{K} \][/tex]
The relationship between [tex]\( K_c \)[/tex] and [tex]\( K_p \)[/tex] is given by the equation:
[tex]\[ K_p = K_c (RT)^{\Delta n} \][/tex]
Where:
- [tex]\( R \)[/tex] is the ideal gas constant. Given as [tex]\( R = 0.08206\ \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \)[/tex],
- [tex]\( T \)[/tex] is the temperature in Kelvin, which we've found to be [tex]\( 1546.15\ \text{K} \)[/tex],
- [tex]\( \Delta n \)[/tex] is the change in the number of moles of gas.
Next, we compute [tex]\( \Delta n \)[/tex], which is the difference between the number of moles of gaseous products and the number of moles of gaseous reactants. Specifically:
[tex]\[ \Delta n = (\text{moles of CO}_2) - (\text{moles of CO} + \text{moles of O}_2) \][/tex]
From the balanced equation:
- The number of moles of products (CO[tex]\(_2\)[/tex]) is 2.
- The number of moles of reactants (CO + O[tex]\(_2\)[/tex]) is 2 + 1 = 3.
Thus:
[tex]\[ \Delta n = 2 - 3 = -1 \][/tex]
Now we can substitute all known values into the [tex]\( K_p \)[/tex] formula:
[tex]\[ K_p = K_c \cdot (RT)^{\Delta n} = 2.24 \times 10^{22} \cdot (0.08206 \cdot 1546.15)^{-1} \][/tex]
To proceed further:
[tex]\[ RT = 0.08206 \cdot 1546.15 = 126.89749 \][/tex]
Hence:
[tex]\[ (RT)^{\Delta n} = (126.89749)^{-1} = \frac{1}{126.89749} \approx 0.007879 \][/tex]
Now, substitute [tex]\( (RT)^{\Delta n} \)[/tex] back into the equation for [tex]\( K_p \)[/tex]:
[tex]\[ K_p \approx 2.24 \times 10^{22} \cdot 0.007879 \][/tex]
Therefore:
[tex]\[ K_p \approx 1.7654884508720805 \times 10^{20} \][/tex]
So, the equilibrium constant [tex]\( K_p \)[/tex] at [tex]\( 1273.0^\circ \text{C} \)[/tex] is approximately:
[tex]\[ K_p \approx 1.77 \times 10^{20} \][/tex]
This concludes our calculation.
