At a certain temperature, the equilibrium constant `K_(c)` is 16 for the reaction,
`SO_((g))+NO_(2(g))hArrSO_(3(g))+NO_((g))`
If 1.0 mol each of the four gases is taken in a one litre container the concentration of `NO_(2)` at equilibrium would is

Options

A. 1.6 mol `L^(-1)`
B. 0.8 mol `L^(-1)`
C. 0.4 mol `L^-1`
D. 0.6 mol `L^(-1)`

Answer

To find the concentration of \( NO_2 \) at equilibrium for the reaction:

\[ SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g) \]

given that the equilibrium constant \( K_c \) is 16, we can follow these steps:

### Step 1: Set up the initial concentrations

Initially, we have 1 mole of each gas in a 1-liter container. Therefore, the initial concentrations are:

\[ [SO_2] = 1 \, \text{M}, \quad [NO_2] = 1 \, \text{M}, \quad [SO_3] = 1 \, \text{M}, \quad [NO] = 1 \, \text{M} \]

### Step 2: Define the change in concentration

Let \( \alpha \) be the amount of \( SO_2 \) and \( NO_2 \) that react at equilibrium. The changes in concentration will be:

\[ [SO_2] = 1 - \alpha \\ [NO_2] = 1 - \alpha \\ [SO_3] = 1 + \alpha \\ [NO] = 1 + \alpha \]

### Step 3: Write the expression for the equilibrium constant

The expression for the equilibrium constant \( K_c \) is given by:

\[ K_c = \frac{[SO_3][NO]}{[SO_2][NO_2]} \]

Substituting the equilibrium concentrations into the expression, we have:

\[ K_c = \frac{(1 + \alpha)(1 + \alpha)}{(1 - \alpha)(1 - \alpha)} = \frac{(1 + \alpha)^2}{(1 - \alpha)^2} \]

### Step 4: Substitute the value of \( K_c \)

Given that \( K_c = 16 \), we can set up the equation:

\[ 16 = \frac{(1 + \alpha)^2}{(1 - \alpha)^2} \]

### Step 5: Take the square root of both sides

Taking the square root of both sides gives:

\[ 4 = \frac{1 + \alpha}{1 - \alpha} \]

### Step 6: Cross-multiply to solve for \( \alpha \)

Cross-multiplying gives:

\[ 4(1 - \alpha) = 1 + \alpha \]

Expanding this, we have:

\[ 4 - 4\alpha = 1 + \alpha \]

### Step 7: Rearrange the equation

Rearranging the equation leads to:

\[ 4 - 1 = 4\alpha + \alpha \\ 3 = 5\alpha \]

### Step 8: Solve for \( \alpha \)

Solving for \( \alpha \):

\[ \alpha = \frac{3}{5} = 0.6 \]

### Step 9: Calculate the equilibrium concentration of \( NO_2 \)

Now, we can find the equilibrium concentration of \( NO_2 \):

\[ [NO_2] = 1 - \alpha = 1 - 0.6 = 0.4 \, \text{M} \]

### Final Answer

The concentration of \( NO_2 \) at equilibrium is:

\[ \boxed{0.4 \, \text{M}} \]

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