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For the reaction 2SO(3)(g)hArr2SO(2(g))...

For the reaction `2SO_(3)(g)hArr2SO_(2(g))+O_(2(g)) ` `(K_(C))/(K_(p))` will be :-

A

`(RT)^(1)`

B

`(RT)^(-1)`

C

`sqrt(RT)`

D

`(RT)^(-2)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the relationship between \( K_c \) and \( K_p \) for the reaction \[ 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \] we will follow these steps: ### Step 1: Identify the reaction and the equilibrium constants We have the reaction: \[ 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \] We need to find the ratio \( \frac{K_c}{K_p} \). ### Step 2: Use the relationship between \( K_p \) and \( K_c \) The relationship between \( K_p \) and \( K_c \) is given by the formula: \[ K_p = K_c (RT)^{\Delta N_g} \] where: - \( R \) is the universal gas constant, - \( T \) is the temperature in Kelvin, - \( \Delta N_g \) is the change in the number of moles of gas. ### Step 3: Calculate \( \Delta N_g \) To find \( \Delta N_g \), we use the formula: \[ \Delta N_g = N_p - N_r \] where: - \( N_p \) is the total number of moles of gaseous products, - \( N_r \) is the total number of moles of gaseous reactants. From the reaction: - Products: \( 2SO_2 + 1O_2 \) gives \( N_p = 2 + 1 = 3 \) - Reactants: \( 2SO_3 \) gives \( N_r = 2 \) Thus, \[ \Delta N_g = 3 - 2 = 1 \] ### Step 4: Substitute \( \Delta N_g \) into the equation for \( K_p \) Now we can substitute \( \Delta N_g \) into the equation for \( K_p \): \[ K_p = K_c (RT)^{1} \] This simplifies to: \[ K_p = K_c RT \] ### Step 5: Rearrange to find \( \frac{K_c}{K_p} \) To find \( \frac{K_c}{K_p} \), we rearrange the equation: \[ K_c = \frac{K_p}{RT} \] Thus, \[ \frac{K_c}{K_p} = \frac{1}{RT} \] ### Final Result Therefore, the final expression for \( \frac{K_c}{K_p} \) is: \[ \frac{K_c}{K_p} = \frac{1}{RT} \]
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