Heats of formation of `SiO_(2)` and `MgO` are `-48.24` and `-34.7` kJ respectively. The heat of the reaction : `2Mg+SiO_(2) rarr 2MgO+Si` is

Options

A. `21.64 kJ`
B. `-21.16 kJ`
C. `-13.62 kJ`
D. `13.62 kJ`

Explanation

To find the heat of the reaction for the equation \(2Mg + SiO_2 \rightarrow 2MgO + Si\), we will use the heats of formation of the reactants and products involved in the reaction. The heats of formation for \(SiO_2\) and \(MgO\) are given as follows:

- Heat of formation of \(SiO_2\) = \(-48.24 \, \text{kJ/mol}\)
- Heat of formation of \(MgO\) = \(-34.7 \, \text{kJ/mol}\)

### Step-by-Step Solution:

1. **Write the formation reactions:**
- For \(SiO_2\):
\[ Si + O_2 \rightarrow SiO_2 \quad \Delta H = -48.24 \, \text{kJ} \]
- For \(MgO\):
\[ Mg + \frac{1}{2} O_2 \rightarrow MgO \quad \Delta H = -34.7 \, \text{kJ} \]

2. **Reverse the reaction for \(SiO_2\) to match the given reaction:**
- Reversing the formation reaction gives:
\[ SiO_2 \rightarrow Si + O_2 \quad \Delta H = +48.24 \, \text{kJ} \]

3. **Multiply the formation reaction of \(MgO\) by 2:**
- This gives:
\[ 2Mg + O_2 \rightarrow 2MgO \quad \Delta H = 2 \times (-34.7) = -69.4 \, \text{kJ} \]

4. **Combine the two reactions:**
- Adding the reversed reaction for \(SiO_2\) and the multiplied reaction for \(MgO\):
\[ SiO_2 \rightarrow Si + O_2 \quad (+48.24 \, \text{kJ}) \\ 2Mg + O_2 \rightarrow 2MgO \quad (-69.4 \, \text{kJ}) \]
- This results in:
\[ 2Mg + SiO_2 \rightarrow 2MgO + Si \]

5. **Calculate the total heat of the reaction:**
- The total heat change (\(\Delta H\)) for the reaction is:
\[ \Delta H = (-69.4) + (+48.24) = -21.16 \, \text{kJ} \]

6. **Final Answer:**
- The heat of the reaction \(2Mg + SiO_2 \rightarrow 2MgO + Si\) is approximately \(-21.16 \, \text{kJ}\).