Given the reaction between 2 gases represented by `A_(2)` and `B_(2)` to given the compound AB(g). `A_(2)(g) + B_(2)(g)hArr 2AB(g)`
At equilibrium, the concentrtation
of `A_(2) = 3.0xx10^(-3) M`
of `B_(2) = 4.2xx10^(-3) M`
of `AB = 2.8xx10^(-3) M`
If the reaction takes place in a sealed vessel at `527^(@)C` . then the value of `K_(c)` will be

Options

A. `2.0`
B. `1.9`
C. `0.62`
D. `4.5`

Answer

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

\[ A_2(g) + B_2(g) \rightleftharpoons 2AB(g) \]

we will use the equilibrium concentrations provided:

- Concentration of \( A_2 = 3.0 \times 10^{-3} \, M \)
- Concentration of \( B_2 = 4.2 \times 10^{-3} \, M \)
- Concentration of \( AB = 2.8 \times 10^{-3} \, M \)

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

The equilibrium constant \( K_c \) is defined as:

\[ K_c = \frac{[\text{products}]^{\text{coefficients}}}{[\text{reactants}]^{\text{coefficients}}} \]

For our reaction, this becomes:

\[ K_c = \frac{[AB]^2}{[A_2][B_2]} \]

### Step 2: Substitute the equilibrium concentrations into the expression

Now we will substitute the equilibrium concentrations into the expression for \( K_c \):

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

### Step 3: Calculate the numerator

First, calculate the numerator:

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

### Step 4: Calculate the denominator

Next, calculate the denominator:

\[ (3.0 \times 10^{-3})(4.2 \times 10^{-3}) = 1.26 \times 10^{-5} \]

### Step 5: Divide the numerator by the denominator

Now we can calculate \( K_c \):

\[ K_c = \frac{7.84 \times 10^{-6}}{1.26 \times 10^{-5}} \approx 0.62 \]

### Final Answer

Thus, the value of \( K_c \) is approximately:

\[ K_c \approx 0.62 \]

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