The equilibrium `K_(c)`for the reaction `SO_(2)(g)NO_(2)(g)hArrSO_(3)(g)+NO(g)is 16` 1 mole of rach of all the four gases is taken in `1dm^(3)` vessel , the equilibrium concentration of NO would be:

Options

A. 0.4M
B. 0.6M
C. 1.4M
D. 1.6M

Answer

To solve the problem, we need to determine the equilibrium concentration of NO for the given reaction:

\[ \text{SO}_2(g) + \text{NO}_2(g) \rightleftharpoons \text{SO}_3(g) + \text{NO}(g) \]

Given that the equilibrium constant \( K_c \) for this reaction is 16, and we start with 1 mole of each gas in a 1 dm³ vessel, we can follow these steps:

### Step-by-Step Solution:

1. **Initial Concentrations**:
Since we have 1 mole of each gas in a 1 dm³ vessel, the initial concentrations of all gases are:
\[ [\text{SO}_2] = 1 \, \text{M}, \quad [\text{NO}_2] = 1 \, \text{M}, \quad [\text{SO}_3] = 1 \, \text{M}, \quad [\text{NO}] = 1 \, \text{M} \]

2. **Calculate Reaction Quotient \( Q_c \)**:
The reaction quotient \( Q_c \) is given by the expression:
\[ Q_c = \frac{[\text{SO}_3][\text{NO}]}{[\text{SO}_2][\text{NO}_2]} \]
Substituting the initial concentrations:
\[ Q_c = \frac{(1)(1)}{(1)(1)} = 1 \]

3. **Compare \( Q_c \) with \( K_c \)**:
Since \( K_c = 16 \) and \( Q_c = 1 \), we find that:
\[ Q_c < K_c \]
This indicates that the reaction will proceed in the forward direction to reach equilibrium.

4. **Change in Concentrations**:
Let \( x \) be the amount of SO₂ and NO₂ that reacts to reach equilibrium. Therefore, at equilibrium:
\[ [\text{SO}_2] = 1 - x, \quad [\text{NO}_2] = 1 - x, \quad [\text{SO}_3] = 1 + x, \quad [\text{NO}] = 1 + x \]

5. **Set Up the Equilibrium Expression**:
The equilibrium constant expression becomes:
\[ K_c = \frac{[\text{SO}_3][\text{NO}]}{[\text{SO}_2][\text{NO}_2]} = \frac{(1 + x)(1 + x)}{(1 - x)(1 - x)} = 16 \]
Simplifying this gives:
\[ K_c = \frac{(1 + x)^2}{(1 - x)^2} = 16 \]

6. **Cross Multiply and Simplify**:
Cross multiplying gives:
\[ (1 + x)^2 = 16(1 - x)^2 \]
Expanding both sides:
\[ 1 + 2x + x^2 = 16(1 - 2x + x^2) \]
\[ 1 + 2x + x^2 = 16 - 32x + 16x^2 \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ 0 = 15x^2 - 34x + 15 \]

8. **Using the Quadratic Formula**:
We can solve for \( x \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{34 \pm \sqrt{(-34)^2 - 4 \cdot 15 \cdot 15}}{2 \cdot 15} \]
\[ x = \frac{34 \pm \sqrt{1156 - 900}}{30} = \frac{34 \pm \sqrt{256}}{30} = \frac{34 \pm 16}{30} \]
This gives us two possible values for \( x \):
\[ x = \frac{50}{30} = \frac{5}{3} \quad \text{or} \quad x = \frac{18}{30} = \frac{3}{5} \]

9. **Selecting the Valid Solution**:
Since \( x \) must be less than 1 (as it represents the amount reacted), we take:
\[ x = \frac{3}{5} \]

10. **Calculating the Equilibrium Concentration of NO**:
The equilibrium concentration of NO is:
\[ [\text{NO}] = 1 + x = 1 + \frac{3}{5} = \frac{8}{5} = 1.6 \, \text{M} \]

### Final Answer:
The equilibrium concentration of NO is \( 1.6 \, \text{M} \).