The correct order of equilibrium constants for the reactions is `H_(2)CO+H_(2)Ooverset(K_1)hArrH_(2)C(OH)_2`
`CH_3CH_2CHO+H_2Ooverset(K_2)hArrCH_3CH_2CH(OH)_2`
`CH_3COCH_3+H_2Ooverset(K_3)hArrCH_3C(OH)_(2)CH_3`

Options

A. `K_1 gt K_2 gt K_3`
B. `K_1 lt K_2 lt K_3`
C. `K_1 gt K_3 gt K_2`
D. `K_1 lt K_3 lt K_2`

Explanation

To determine the correct order of equilibrium constants \( K_1 \), \( K_2 \), and \( K_3 \) for the given reactions, we need to analyze the structure of the reactants and the steric effects involved in each reaction.

### Step 1: Identify the Reactions
The reactions given are:
1. \( \text{H}_2\text{CO} + \text{H}_2\text{O} \overset{K_1}{\rightleftharpoons} \text{H}_2\text{C(OH)}_2 \)
2. \( \text{CH}_3\text{CH}_2\text{CHO} + \text{H}_2\text{O} \overset{K_2}{\rightleftharpoons} \text{CH}_3\text{CH}_2\text{CH(OH)}_2 \)
3. \( \text{CH}_3\text{COCH}_3 + \text{H}_2\text{O} \overset{K_3}{\rightleftharpoons} \text{CH}_3\text{C(OH)}_2\text{CH}_3 \)

### Step 2: Analyze the Reactants
- **Reaction 1** involves formaldehyde (\( \text{H}_2\text{CO} \)), which has no bulky groups attached to the carbonyl carbon.
- **Reaction 2** involves an aldehyde with one alkyl group (\( \text{CH}_3\text{CH}_2\text{CHO} \)), which introduces some steric hindrance.
- **Reaction 3** involves a ketone with two alkyl groups (\( \text{CH}_3\text{COCH}_3 \)), which introduces more steric hindrance compared to the previous reactions.

### Step 3: Consider Steric Effects
Steric hindrance affects the reactivity of the carbonyl group towards nucleophilic addition (in this case, the addition of water). The more bulky groups attached to the carbonyl carbon, the greater the steric hindrance, which will decrease the equilibrium constant for the hydration reaction.

### Step 4: Determine the Order of Equilibrium Constants
- **For \( K_1 \)**: The least steric hindrance (no bulky groups) means the equilibrium constant will be the highest.
- **For \( K_2 \)**: One bulky group leads to moderate steric hindrance, resulting in a lower equilibrium constant than \( K_1 \).
- **For \( K_3 \)**: Two bulky groups lead to the highest steric hindrance, resulting in the lowest equilibrium constant.

Thus, the order of equilibrium constants is:
\[ K_1 > K_2 > K_3 \]

### Final Answer
The correct order of equilibrium constants is:
\[ K_1 < K_2 < K_3 \]