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

Explanation

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 in the question.

### Step-by-Step Solution:

1. **Write the expression for \( K_c \)**:
The equilibrium constant \( K_c \) for the reaction can be expressed as:

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

2. **Substitute the equilibrium concentrations**:
From the problem, we have:
- \([A_2] = 3.0 \times 10^{-3} \, M\)
- \([B_2] = 4.2 \times 10^{-3} \, M\)
- \([AB] = 2.8 \times 10^{-3} \, M\)

Plugging these values into the \( K_c \) expression:

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

3. **Calculate the numerator**:
Calculate \((2.8 \times 10^{-3})^2\):

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

4. **Calculate the denominator**:
Calculate \((3.0 \times 10^{-3})(4.2 \times 10^{-3})\):

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

5. **Combine the results**:
Now substitute the results back into the \( K_c \) expression:

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

6. **Perform the division**:
Calculate \( \frac{7.84}{12.6} \):

\[ K_c \approx 0.622 \]

Rounding to two decimal places gives:

\[ K_c \approx 0.62 \]

### Final Answer:
Thus, the value of \( K_c \) is approximately **0.62**.

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