Determine the value of Kp for the oxidation of sulfur dioxide at 298 K given that K. = 7.0 x 1024. 2 SO2(g) + O2(g) = 2 SO3(g) Kp = K.(TR)An where R = 0.08216 L.atm/mol.K Kp =
Determine the value of Kp for the oxidation of sulfur dioxide at 298 K given that K. = 7.0 x 1024. 2 SO2(g) + O2(g) = 2 SO3(g) Kp = K.(TR)An where R = 0.08216 L.atm/mol.K Kp =
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![### Determining the Value of \( K_p \) for the Oxidation of Sulfur Dioxide
Given the chemical equation for the oxidation of sulfur dioxide at 298 K:
\[ 2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g) \]
We are provided with the equilibrium constant in terms of concentration (\( K_c \)):
\[ K_c = 7.0 \times 10^{24} \]
To find the equilibrium constant in terms of pressure (\( K_p \)), we use the following formula:
\[ K_p = K_c (RT)^{\Delta n} \]
where:
- \( R \) is the ideal gas constant, which equals \( 0.08216 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \).
- \( T \) is the temperature in Kelvin, which is 298 K in this case.
- \( \Delta n \) is the change in moles of gas, calculated as follows:
\[ \Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) \]
For this reaction:
\[ \Delta n = (2) - (2 + 1) = 2 - 3 = -1 \]
With \( \Delta n = -1 \), the equation becomes:
\[ K_p = K_c (RT)^{-1} \]
Finally, by inserting the values into the equation, we can determine \( K_p \):
\[ K_p = 7.0 \times 10^{24} \left( 0.08216 \cdot 298 \right)^{-1} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F48deb947-3d41-43de-ac43-bef03e01cc59%2F2a223421-3d56-4aae-bf05-c5a4654c721a%2Fiycwz5i.png&w=3840&q=75)
Transcribed Image Text:### Determining the Value of \( K_p \) for the Oxidation of Sulfur Dioxide
Given the chemical equation for the oxidation of sulfur dioxide at 298 K:
\[ 2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g) \]
We are provided with the equilibrium constant in terms of concentration (\( K_c \)):
\[ K_c = 7.0 \times 10^{24} \]
To find the equilibrium constant in terms of pressure (\( K_p \)), we use the following formula:
\[ K_p = K_c (RT)^{\Delta n} \]
where:
- \( R \) is the ideal gas constant, which equals \( 0.08216 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \).
- \( T \) is the temperature in Kelvin, which is 298 K in this case.
- \( \Delta n \) is the change in moles of gas, calculated as follows:
\[ \Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) \]
For this reaction:
\[ \Delta n = (2) - (2 + 1) = 2 - 3 = -1 \]
With \( \Delta n = -1 \), the equation becomes:
\[ K_p = K_c (RT)^{-1} \]
Finally, by inserting the values into the equation, we can determine \( K_p \):
\[ K_p = 7.0 \times 10^{24} \left( 0.08216 \cdot 298 \right)^{-1} \]
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