At 350 K, `K_(p)` for the reaction given below is `3.0xx10^(10)"bar"^(-1)` at equilibrium. What be the value of `K_(c)` at this temperature ?
`2N_(2(g))+O_(2(g))hArr2N_(2)O_((g))`

Options

A. `7.4xx10^(11)"L mol"^(-1)`
B. `8715xx10^(10)"L mol"^(-1)`
C. `0.08"L mol"^(-1)`
D. `8.715xx10^(11)"L mol"^(-1)`

Explanation

To find the value of \( K_c \) at 350 K for the reaction:

\[ 2N_{2(g)} + O_{2(g)} \rightleftharpoons 2N_{2}O_{(g)} \]

we can use the relationship between \( K_p \) and \( K_c \):

\[ K_p = K_c \cdot R T^{\Delta n} \]

where:
- \( R \) is the ideal gas constant (0.083 L·bar/K·mol),
- \( T \) is the temperature in Kelvin,
- \( \Delta n \) is the change in the number of moles of gas (moles of products - moles of reactants).

### Step 1: Calculate \( \Delta n \)

First, we need to determine \( \Delta n \):

- Moles of gaseous products: \( 2 \) moles of \( N_2O \)
- Moles of gaseous reactants: \( 2 \) moles of \( N_2 \) + \( 1 \) mole of \( O_2 \) = \( 3 \) moles

Now, calculate \( \Delta n \):

\[ \Delta n = \text{moles of products} - \text{moles of reactants} = 2 - 3 = -1 \]

### Step 2: Rearrange the equation for \( K_c \)

Now, we can rearrange the equation to solve for \( K_c \):

\[ K_c = \frac{K_p}{R T^{\Delta n}} \]

### Step 3: Substitute the known values

Given:
- \( K_p = 3.0 \times 10^{10} \, \text{bar}^{-1} \)
- \( R = 0.083 \, \text{L·bar/K·mol} \)
- \( T = 350 \, \text{K} \)
- \( \Delta n = -1 \)

Substituting these values into the equation:

\[ K_c = \frac{3.0 \times 10^{10}}{0.083 \times 350^{-1}} \]

Since \( \Delta n = -1 \), we can rewrite \( T^{\Delta n} \) as \( \frac{1}{T} \):

\[ K_c = 3.0 \times 10^{10} \times (0.083 \times 350) \]

### Step 4: Calculate \( K_c \)

Now calculate \( K_c \):

\[ K_c = 3.0 \times 10^{10} \times (0.083 \times 350) \]

Calculating \( 0.083 \times 350 \):

\[ 0.083 \times 350 = 29.05 \]

Now substituting back:

\[ K_c = 3.0 \times 10^{10} \times 29.05 \]

Calculating \( K_c \):

\[ K_c = 8.715 \times 10^{11} \, \text{L/mol} \]

### Final Answer

Thus, the value of \( K_c \) at 350 K is:

\[ K_c \approx 8.715 \times 10^{11} \, \text{L/mol} \]

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