At `460^(@)C, K_(C)=81` for the reaction, `SO_(2)(g)+NO_(2)(g)hArrNO(g)+SO_(3)(g)`
A mixture of these gases has the following concentrations of the reactants and products:
`[SO_(2)]=0.04M [NO_(2)]=0.04m`
`[NO]-0.30m [SO]=0.3m`
Is the system at equilibrium? If not, in which direction must the reaction proceed to reach equilibrium. What will be the molar concentrations of the four gases at equilibrium?

Explanation

To determine whether the system is at equilibrium and to find the equilibrium concentrations of the gases involved in the reaction \( SO_2(g) + NO_2(g) \rightleftharpoons NO(g) + SO_3(g) \), we will follow these steps:

### Step 1: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant expression for the reaction is given by:

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

### Step 2: Identify the initial concentrations
From the problem, we have the following initial concentrations:

- \([SO_2] = 0.04 \, M\)
- \([NO_2] = 0.04 \, M\)
- \([NO] = 0.30 \, M\)
- \([SO_3] = 0.30 \, M\)

### Step 3: Calculate the reaction quotient \( Q_c \)
We can calculate the reaction quotient \( Q_c \) using the initial concentrations:

\[ Q_c = \frac{[NO][SO_3]}{[SO_2][NO_2]} = \frac{(0.30)(0.30)}{(0.04)(0.04)} = \frac{0.09}{0.0016} = 56.25 \]

### Step 4: Compare \( Q_c \) with \( K_c \)
Given that \( K_c = 81 \), we compare \( Q_c \) with \( K_c \):

- \( Q_c = 56.25 \)
- \( K_c = 81 \)

Since \( Q_c < K_c \), the system is not at equilibrium.

### Step 5: Determine the direction of the reaction
Since \( Q_c < K_c \), the reaction will proceed in the forward direction (to the right) to reach equilibrium.

### Step 6: Set up the equilibrium concentrations
Let \( x \) be the change in concentration of the reactants and products at equilibrium. The changes will be:

- \([SO_2] = 0.04 - x\)
- \([NO_2] = 0.04 - x\)
- \([NO] = 0.30 + x\)
- \([SO_3] = 0.30 + x\)

### Step 7: Write the equilibrium expression with \( x \)
At equilibrium, we can express \( K_c \) as:

\[ K_c = \frac{(0.30 + x)(0.30 + x)}{(0.04 - x)(0.04 - x)} = 81 \]

### Step 8: Solve for \( x \)
Expanding the equation:

\[ (0.30 + x)^2 = 81 \cdot (0.04 - x)^2 \]

Calculating both sides, we can simplify and solve for \( x \):

1. Expand both sides.
2. Rearrange to form a quadratic equation.
3. Solve for \( x \).

After solving, we find \( x = 0.006 \).

### Step 9: Calculate the equilibrium concentrations
Now we can substitute \( x \) back into the expressions for the equilibrium concentrations:

- \([SO_2] = 0.04 - 0.006 = 0.034 \, M\)
- \([NO_2] = 0.04 - 0.006 = 0.034 \, M\)
- \([NO] = 0.30 + 0.006 = 0.306 \, M\)
- \([SO_3] = 0.30 + 0.006 = 0.306 \, M\)

### Final Answer
The molar concentrations at equilibrium are:
- \([SO_2] = 0.034 \, M\)
- \([NO_2] = 0.034 \, M\)
- \([NO] = 0.306 \, M\)
- \([SO_3] = 0.306 \, M\)