When KBr is treated with conc. `H_(2)SO_(4)` reddish-brown gas is evolved. The gas is 

Options

A. `Br_(2)`
B. `HBr`
C. `NO_(2)`
D. `H_(2)O_(2)`

Answer

To solve the question regarding the reaction of KBr with concentrated H₂SO₄ and the reddish-brown gas evolved, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants in this reaction are potassium bromide (KBr) and concentrated sulfuric acid (H₂SO₄).

2. **Understand the Nature of KBr**:
- KBr is a salt formed from a strong base (KOH) and a weak acid (HBr). When salts are treated with concentrated acids, they can undergo a reaction that produces gases.

3. **Reaction with Concentrated H₂SO₄**:
- When KBr reacts with concentrated H₂SO₄, it can lead to the release of bromine gas (Br₂) along with the formation of potassium sulfate (K₂SO₄).

4. **Write the Balanced Reaction**:
- The balanced chemical equation for the reaction can be represented as:
\[ 2 KBr + H₂SO₄ \rightarrow K₂SO₄ + 2 HBr \]
- However, since HBr is a weak acid, it can further react with concentrated H₂SO₄ to produce bromine gas:
\[ 2 HBr + H₂SO₄ \rightarrow Br₂ + SO₂ + 2 H₂O \]

5. **Identify the Gas Produced**:
- The reddish-brown gas evolved in this reaction is bromine (Br₂).

6. **Conclusion**:
- Therefore, the reddish-brown gas evolved when KBr is treated with concentrated H₂SO₄ is bromine (Br₂).

### Final Answer:
The reddish-brown gas evolved is **Br₂ (bromine)**.

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