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 find the equilibrium concentration of NO for the 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 initially, we have 1 mole of each gas in a 1 dm³ vessel.

### Step-by-Step Solution:

1. **Write the Reaction and Initial Conditions:**
The balanced chemical equation is:
\[ \text{SO}_2(g) + \text{NO}_2(g) \rightleftharpoons \text{SO}_3(g) + \text{NO}(g) \]
Initial moles of each gas:
\[ [\text{SO}_2] = 1 \, \text{mol}, \quad [\text{NO}_2] = 1 \, \text{mol}, \quad [\text{SO}_3] = 1 \, \text{mol}, \quad [\text{NO}] = 1 \, \text{mol} \]

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

3. **Write the Expression for \( K_c \):**
The expression for the equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[\text{SO}_3][\text{NO}]}{[\text{SO}_2][\text{NO}_2]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(1 + x)(1 + x)}{(1 - x)(1 - x)} = \frac{(1 + x)^2}{(1 - x)^2} \]

4. **Substitute the Given \( K_c \) Value:**
We know \( K_c = 16 \):
\[ 16 = \frac{(1 + x)^2}{(1 - x)^2} \]

5. **Cross-Multiply and Simplify:**
\[ 16(1 - x)^2 = (1 + x)^2 \]
Expanding both sides:
\[ 16(1 - 2x + x^2) = 1 + 2x + x^2 \]
\[ 16 - 32x + 16x^2 = 1 + 2x + x^2 \]
Rearranging gives:
\[ 15x^2 - 34x + 15 = 0 \]

6. **Solve the Quadratic Equation:**
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = 15, \, b = -34, \, c = 15 \]
\[ x = \frac{34 \pm \sqrt{(-34)^2 - 4 \cdot 15 \cdot 15}}{2 \cdot 15} \]
\[ x = \frac{34 \pm \sqrt{1156 - 900}}{30} \]
\[ x = \frac{34 \pm \sqrt{256}}{30} \]
\[ x = \frac{34 \pm 16}{30} \]
This gives two possible values for \( x \):
\[ x = \frac{50}{30} = \frac{5}{3} \quad \text{(not feasible as it exceeds initial moles)} \]
\[ x = \frac{18}{30} = \frac{3}{5} \]

7. **Calculate the Equilibrium Concentration of NO:**
The equilibrium concentration of NO is:
\[ [\text{NO}] = 1 + x = 1 + \frac{3}{5} = \frac{5}{5} + \frac{3}{5} = \frac{8}{5} = 1.6 \, \text{mol/dm}^3 \]

### Final Answer:
The equilibrium concentration of NO is \( 1.6 \, \text{mol/dm}^3 \).

---