In which of the following does the reaction go almost to completion ?

Options

A. `K_c=10^3`
B. `K_=c=10^2`
C. `K_c=10^(-2)`
D. `K_c=10^(-3)`

Answer

To determine in which of the given reactions the reaction goes almost to completion, we need to analyze the equilibrium constant (Kc) values provided for each reaction. The Kc value indicates the extent of a reaction at equilibrium, with higher values suggesting that the reaction favors the formation of products.

### Step-by-Step Solution:

1. **Understanding Kc**: The equilibrium constant (Kc) is defined for a reaction at equilibrium as:
\[ K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} \]
where \([C]\) and \([D]\) are the concentrations of the products, and \([A]\) and \([B]\) are the concentrations of the reactants, with \(a\), \(b\), \(c\), and \(d\) being their respective stoichiometric coefficients.

2. **Interpreting Kc Values**: A higher Kc value indicates that at equilibrium, the concentration of products is much greater than that of reactants. Conversely, a lower Kc value indicates that the reactants are favored.

3. **Given Kc Values**: We are provided with four different Kc values:
- Option 1: \(K_c = 10^3\)
- Option 2: \(K_c = 10^2\)
- Option 3: \(K_c = 10^{-2}\)
- Option 4: \(K_c = 10^{-3}\)

4. **Comparing Kc Values**: To find out which reaction goes almost to completion, we compare the Kc values:
- \(10^3\) (Option 1) is the highest.
- \(10^2\) (Option 2) is lower than \(10^3\).
- \(10^{-2}\) (Option 3) is significantly lower.
- \(10^{-3}\) (Option 4) is the lowest.

5. **Conclusion**: Since the highest Kc value is \(10^3\) from Option 1, we conclude that the reaction corresponding to this Kc value goes almost to completion. This means that at equilibrium, the concentration of products is significantly higher than that of reactants.

### Final Answer:
The reaction that goes almost to completion is the one with \(K_c = 10^3\) (Option 1).