One of the greenhouse gas is

Options

A. CO
B. `O_(2)`
C. `H_(2)`
D. CFC's

Answer

To solve the question regarding the equivalent weight of H₂SO₄ in the given reaction, we will follow these steps:

### Step 1: Identify the Reaction
First, we need to identify the reaction in which H₂SO₄ is involved. In this case, it is likely a redox reaction where sulfur dioxide (SO₂) is oxidized to sulfate (SO₄²⁻) and dichromate (Cr₂O₇²⁻) is reduced.

### Step 2: Determine Oxidation States
Next, we will determine the oxidation states of sulfur in SO₂ and SO₄²⁻, as well as chromium in Cr₂O₇²⁻.

- For SO₂: The oxidation state of sulfur (S) is +4.
- For SO₄²⁻: The oxidation state of sulfur (S) is +6.
- For Cr₂O₇²⁻: The oxidation state of chromium (Cr) is +6.

### Step 3: Calculate the Change in Oxidation States
Now, we calculate the change in oxidation states for sulfur and chromium:

- Sulfur: Changes from +4 to +6, which means it loses 2 electrons (oxidation).
- Chromium: Changes from +6 to +3, which means it gains 3 electrons (reduction).

### Step 4: Balance the Electrons
To balance the electrons in the half-reactions, we need to make sure that the total number of electrons lost equals the total number of electrons gained. Since sulfur loses 2 electrons and chromium gains 3 electrons, we can multiply the sulfur half-reaction by 3 to balance:

- 3(SO₂ → SO₄²⁻) will involve 6 electrons.
- Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ will also involve 6 electrons.

### Step 5: Calculate the Equivalent Weight
The equivalent weight of H₂SO₄ can be calculated using the formula:

\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{n \text{ factor}} \]

Where:
- Molar Mass of H₂SO₄ = 2 (H) + 32 (S) + 4 × 16 (O) = 98 g/mol
- n factor = 6 (since 6 electrons are involved in the reaction).

Now, substituting the values:

\[ \text{Equivalent Weight of H₂SO₄} = \frac{98 \text{ g/mol}}{6} = \frac{98}{6} \text{ g/equiv} \]

### Final Answer
The equivalent weight of H₂SO₄ in the reaction is \(\frac{98}{6}\) g/equiv.

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