Givecn that :
`Zn+1//2O_(2)rarr ZnO+84000` cal ……………1
`Hg+1//2O_(2)rarr HgO+21700` cal …………2
The heat of reaction `(Delta H)` for,
`Zn+HgO rarr ZnO+Hg` is :-

Options

A. 105700 cal
B. 62300 cal
C. `-105700` cal
D. `-62300` cal

Answer

To find the heat of reaction (ΔH) for the reaction:

\[ \text{Zn} + \text{HgO} \rightarrow \text{ZnO} + \text{Hg} \]

we can use Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction.

### Step-by-Step Solution:

1. **Write down the given reactions and their enthalpy changes:**
- Reaction 1:
\[ \text{Zn} + \frac{1}{2} \text{O}_2 \rightarrow \text{ZnO} + 84000 \, \text{cal} \]
- Reaction 2:
\[ \text{Hg} + \frac{1}{2} \text{O}_2 \rightarrow \text{HgO} + 21700 \, \text{cal} \]

2. **Reverse Reaction 2:**
Since we need HgO to be a reactant in our desired reaction, we reverse Reaction 2:
\[ \text{HgO} \rightarrow \text{Hg} + \frac{1}{2} \text{O}_2 - 21700 \, \text{cal} \]
(Note: The enthalpy change becomes negative when reversing the reaction.)

3. **Combine the reactions:**
Now, we can add Reaction 1 and the reversed Reaction 2:
\[ \text{Zn} + \frac{1}{2} \text{O}_2 \rightarrow \text{ZnO} + 84000 \, \text{cal} \]
\[ \text{HgO} \rightarrow \text{Hg} + \frac{1}{2} \text{O}_2 - 21700 \, \text{cal} \]
When we add these two reactions, the \(\frac{1}{2} \text{O}_2\) cancels out:
\[ \text{Zn} + \text{HgO} \rightarrow \text{ZnO} + \text{Hg} \]

4. **Calculate the total ΔH:**
Now, we can calculate the total ΔH for the overall reaction:
\[ \Delta H = (84000 \, \text{cal}) + (-21700 \, \text{cal}) \]
\[ \Delta H = 84000 - 21700 = 62300 \, \text{cal} \]

5. **Final Result:**
Thus, the heat of reaction (ΔH) for the reaction \( \text{Zn} + \text{HgO} \rightarrow \text{ZnO} + \text{Hg} \) is:
\[ \Delta H = 62300 \, \text{cal} \]