If 200 ml of `0.1 MH_2SO_4` salution is mixed with 200 ml of 0.15 M NaOH solution then heat evolve during the neutralization process is

Options

A. 411 cal
B. 548 cal
C. 959 cal
D. 822 cal

Answer

To solve the problem of calculating the heat evolved during the neutralization of 200 ml of 0.1 M H₂SO₄ with 200 ml of 0.15 M NaOH, we can follow these steps:

### Step 1: Calculate the number of equivalents of H₂SO₄
The normality (N) of H₂SO₄ can be calculated as follows:
- H₂SO₄ is a diprotic acid, meaning it can donate 2 protons (H⁺ ions).
- Therefore, Normality (N) = Molarity (M) × Number of protons = 0.1 M × 2 = 0.2 N

Now, we can calculate the number of equivalents:
- Volume (V) = 200 ml = 0.2 L
- Number of equivalents of H₂SO₄ = N × V = 0.2 N × 0.2 L = 0.04 equivalents (or 40 milliequivalents)

### Step 2: Calculate the number of equivalents of NaOH
For NaOH, which is a monoprotic base:
- Molarity (M) = 0.15 M
- Normality (N) = M = 0.15 N

Now, we can calculate the number of equivalents:
- Volume (V) = 200 ml = 0.2 L
- Number of equivalents of NaOH = N × V = 0.15 N × 0.2 L = 0.03 equivalents (or 30 milliequivalents)

### Step 3: Determine the limiting reagent
From the calculations:
- H₂SO₄ provides 40 milliequivalents.
- NaOH provides 30 milliequivalents.

Since NaOH is the limiting reagent, it will react completely, and there will be some excess H₂SO₄.

### Step 4: Calculate the excess equivalents of H₂SO₄
Excess equivalents of H₂SO₄ = Total equivalents of H₂SO₄ - Total equivalents of NaOH
= 40 milliequivalents - 30 milliequivalents = 10 milliequivalents

### Step 5: Calculate the heat evolved
The heat evolved during the neutralization reaction can be calculated using the heat released per equivalent of reaction:
- Heat released per equivalent = 13.7 kilocalories

Now, we need to calculate the heat evolved for the 30 milliequivalents of NaOH that reacted:
- Heat evolved = (Number of equivalents of NaOH) × (Heat released per equivalent)
= 30 × 13.7 kcal = 411 calories (since 1 kcal = 1000 calories)

### Final Answer
The heat evolved during the neutralization process is **411 calories**.

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