At equilibrium, the concentrations of `N_(2) = 3*0 xx 10^(-3) M, O_(2)= 4*2 xx 10^(-3) M and No = 2*8 xx 10^(-3) M` in a sealed vessel at 800 K. What will be `K_(c)` for the raction
` N_(2) (g) + O_(2) (g) hArr 2 NO (g) ?`

Answer

To find the equilibrium constant \( K_c \) for the reaction

\[ N_2 (g) + O_2 (g) \rightleftharpoons 2 NO (g) \]

we will use the given equilibrium concentrations of the reactants and products.

### Step 1: Write the expression for \( K_c \)

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

\[ K_c = \frac{[NO]^2}{[N_2][O_2]} \]

where:
- \([NO]\) is the concentration of nitrogen monoxide at equilibrium,
- \([N_2]\) is the concentration of nitrogen at equilibrium,
- \([O_2]\) is the concentration of oxygen at equilibrium.

### Step 2: Substitute the given concentrations

From the question, we have the following equilibrium concentrations:
- \([N_2] = 3.0 \times 10^{-3} \, M\)
- \([O_2] = 4.2 \times 10^{-3} \, M\)
- \([NO] = 2.8 \times 10^{-3} \, M\)

Now, we substitute these values into the \( K_c \) expression:

\[ K_c = \frac{(2.8 \times 10^{-3})^2}{(3.0 \times 10^{-3})(4.2 \times 10^{-3})} \]

### Step 3: Calculate the numerator and denominator

First, calculate the numerator:

\[ (2.8 \times 10^{-3})^2 = 7.84 \times 10^{-6} \]

Next, calculate the denominator:

\[ (3.0 \times 10^{-3})(4.2 \times 10^{-3}) = 12.6 \times 10^{-6} \]

### Step 4: Divide the numerator by the denominator

Now, we can calculate \( K_c \):

\[ K_c = \frac{7.84 \times 10^{-6}}{12.6 \times 10^{-6}} \approx 0.622 \]

### Final Answer

Thus, the equilibrium constant \( K_c \) for the reaction at 800 K is approximately:

\[ K_c \approx 0.622 \]

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