For the equilibrium `SO_(2)Cl_(2)(g)hArrSO_(2)(g)+Cl_(2)(g)`, what is the temperature at which`(K_(p)(atm))/(K_(c)(M))=3`?

Options

A. `0.027K `
B. `0.36K`
C. `36.54`K
D. `273K`

Answer

To solve the problem, we need to find the temperature at which the ratio \( \frac{K_p}{K_c} = 3 \) for the equilibrium reaction:

\[ SO_2Cl_2(g) \rightleftharpoons SO_2(g) + Cl_2(g) \]

### Step-by-Step Solution:

1. **Identify the Reaction and Write the Expression for \( K_p \) and \( K_c \)**:
The equilibrium constant \( K_p \) is related to \( K_c \) by the equation:
\[ K_p = K_c \cdot R T^{\Delta n} \]
where:
- \( R \) is the gas constant (0.0821 L·atm/(K·mol))
- \( T \) is the temperature in Kelvin
- \( \Delta n \) is the change in the number of moles of gas, calculated as the moles of gaseous products minus the moles of gaseous reactants.

2. **Calculate \( \Delta n \)**:
For the given reaction:
- Products: \( SO_2(g) + Cl_2(g) \) gives us 2 moles of gas.
- Reactants: \( SO_2Cl_2(g) \) gives us 1 mole of gas.
Thus,
\[ \Delta n = \text{(moles of products)} - \text{(moles of reactants)} = 2 - 1 = 1 \]

3. **Substitute \( \Delta n \) into the Equation**:
Now we can substitute \( \Delta n \) into the equation for \( K_p \):
\[ K_p = K_c \cdot R T^1 = K_c \cdot R T \]

4. **Set Up the Ratio \( \frac{K_p}{K_c} \)**:
Given that \( \frac{K_p}{K_c} = 3 \), we can write:
\[ \frac{K_p}{K_c} = R T = 3 \]

5. **Solve for Temperature \( T \)**:
Rearranging the equation gives us:
\[ T = \frac{3}{R} \]
Substituting \( R = 0.0821 \, \text{L·atm/(K·mol)} \):
\[ T = \frac{3}{0.0821} \approx 36.54 \, \text{K} \]

### Final Answer:
The temperature at which \( \frac{K_p}{K_c} = 3 \) is approximately \( 36.54 \, \text{K} \).

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