M is a metal that forms an oxide `M_(2)O,(1)/(2)M_(2)O rarr M+(1)/(4)O_(2) Delta H=120` K. Cal. When a sample of metal M reacts with one mole of oxygen what will be the `Delta H` in that case

Options

A. 240 K. Cal.
B. `-240` K. Cal.
C. 480 K. Cal.
D. `-480` K. Cal.

Explanation

To solve the problem, we need to analyze the given reaction and the enthalpy change associated with it.

### Step-by-Step Solution:

1. **Understand the Given Reaction:**
The reaction provided is:
\[ \frac{1}{2} M_2O \rightarrow M + \frac{1}{4} O_2 \quad \Delta H = 120 \text{ kcal} \]
This indicates that when half a mole of the metal oxide \( M_2O \) decomposes, it produces 1 mole of metal \( M \) and \( \frac{1}{4} \) mole of oxygen, with an enthalpy change of +120 kcal.

2. **Reverse the Reaction:**
If we reverse the reaction to find the enthalpy change for the formation of \( M_2O \) from \( M \) and \( \frac{1}{4} O_2 \):
\[ M + \frac{1}{4} O_2 \rightarrow \frac{1}{2} M_2O \quad \Delta H = -120 \text{ kcal} \]
This means that the formation of half a mole of \( M_2O \) from \( M \) and \( \frac{1}{4} O_2 \) releases 120 kcal.

3. **Calculate for One Mole of Oxygen:**
We need to find the enthalpy change when 1 mole of oxygen reacts with metal \( M \). Since \( \frac{1}{4} \) mole of \( O_2 \) corresponds to -120 kcal, we can scale this up to 1 mole of \( O_2 \):
\[ \text{For } 1 \text{ mole of } O_2: \quad \Delta H = -120 \text{ kcal} \times 4 = -480 \text{ kcal} \]

4. **Final Result:**
Therefore, when a sample of metal \( M \) reacts with one mole of oxygen, the enthalpy change \( \Delta H \) is:
\[ \Delta H = -480 \text{ kcal} \]

### Conclusion:
The answer to the question is that when a sample of metal \( M \) reacts with one mole of oxygen, the enthalpy change \( \Delta H \) is -480 kcal.