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 determine the enthalpy change (ΔH) when one mole of metal M reacts with oxygen. The given reaction is:

\[ \frac{1}{2} M_2O \rightarrow M + \frac{1}{4} O_2 \quad \Delta H = 120 \text{ kcal} \]

### Step-by-Step Solution:

1. **Understanding the Given Reaction**:
The reaction shows that half a mole of \( M_2O \) decomposes to produce one mole of metal \( M \) and a quarter mole of \( O_2 \) with an enthalpy change of +120 kcal.

2. **Writing the Reaction for One Mole of Metal**:
We need to find the enthalpy change for the reaction where one mole of metal \( M \) reacts with oxygen. The reverse of the given reaction can be written as:
\[ M + \frac{1}{4} O_2 \rightarrow \frac{1}{2} M_2O \]
The enthalpy change for this reverse reaction will be:
\[ \Delta H = -120 \text{ kcal} \]

3. **Scaling the Reaction**:
The reaction we need is for one mole of \( M \) reacting with one mole of \( O_2 \). To find this, we need to scale the reaction. The balanced equation for the formation of \( M_2O \) from \( M \) and \( O_2 \) can be written as:
\[ 2M + O_2 \rightarrow M_2O \]
The enthalpy change for this reaction will be double that of the previous reaction:
\[ \Delta H = 2 \times (-120) \text{ kcal} = -240 \text{ kcal} \]

4. **Finding the ΔH for One Mole of Metal**:
Since the reaction \( 2M + O_2 \rightarrow M_2O \) involves 2 moles of \( M \), the enthalpy change per mole of \( M \) will be:
\[ \Delta H = \frac{-240 \text{ kcal}}{2} = -120 \text{ kcal} \]

### Final Answer:
The enthalpy change (ΔH) when one mole of metal M reacts with one mole of oxygen is:
\[ \Delta H = -120 \text{ kcal} \]