For the reversible reaction, net rate is
`2NO(g)+O_(2)(g) hArr 2NO_(2)(g)`
`((dx)/(dt))_("net")=2.6 xx 10^(3)[NO]^(2)[O_(2)]-4.1 [NO_(2)]^(2)`
If a reaction mixture contains 0.01 mol each of NO and `O_(2)` and 0.1 mol of `NO_(2)` in 1L closed flask, then above reaction is

Options

A. shifted in forward reaction
B. shifted in backward reaction
C. in equilibrium
D. given values are incomplete

Explanation

To determine the nature of the reaction given the concentrations of the reactants and products, we will follow these steps:

### Step 1: Write the expression for the equilibrium constant (Kc)
For the reaction:
\[ 2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g) \]

The equilibrium constant expression is given by:
\[ K_c = \frac{[NO_2]^2}{[NO]^2[O_2]} \]

### Step 2: Calculate Kc using the given rate constants
From the problem, we have:
- Forward rate constant, \( K_f = 2.6 \times 10^3 \)
- Backward rate constant, \( K_b = 4.1 \)

Thus, the equilibrium constant \( K_c \) can be calculated as:
\[ K_c = \frac{K_f}{K_b} = \frac{2.6 \times 10^3}{4.1} \approx 634.15 \approx 6.34 \times 10^2 \]

### Step 3: Determine the concentrations of the reactants and products
Given:
- Concentration of \( NO = 0.01 \, \text{mol/L} \)
- Concentration of \( O_2 = 0.01 \, \text{mol/L} \)
- Concentration of \( NO_2 = 0.1 \, \text{mol/L} \)

### Step 4: Calculate the reaction quotient (Qc)
The reaction quotient \( Q_c \) is calculated using the same expression as \( K_c \):
\[ Q_c = \frac{[NO_2]^2}{[NO]^2[O_2]} = \frac{(0.1)^2}{(0.01)^2 \times (0.01)} = \frac{0.01}{0.0001} = 100 \]

### Step 5: Compare Qc with Kc
Now we compare \( Q_c \) and \( K_c \):
- \( Q_c = 100 \)
- \( K_c \approx 634.15 \)

Since \( Q_c < K_c \), the reaction will shift to the right (forward direction) to reach equilibrium.

### Conclusion
The reaction is currently not at equilibrium, and it will shift to the right (toward the formation of more \( NO_2 \)) to reach equilibrium.

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