When `1L` of `NaOH(1M)` is mixed with `1L` of `HCl(1M)` the temperature of reaction mixture rises by `10^(@)C` . When `1L` of `NaOH(1M)` is mixed with `2L(0.5M)HCl`, the temperature of reaction mixture rises approximately by `:`

Options

A. `10^(@)C`
B. `5^(@)C`
C. `6.67^(@)C`
D. `3.33^(@)C`

Answer

To solve the problem step-by-step, we need to analyze the two scenarios provided in the question.

### Step 1: Understand the first scenario
In the first scenario, we have:
- 1 L of NaOH (1 M) mixed with 1 L of HCl (1 M).
- The temperature rise (ΔT1) is given as 10°C.

Since both solutions are equimolar and of equal volume, we can say that the number of moles of NaOH and HCl are equal.

### Step 2: Calculate the total volume in the first scenario
The total volume (V1) in the first scenario is:
\[ V1 = 1 \, \text{L} + 1 \, \text{L} = 2 \, \text{L} \]

### Step 3: Understand the second scenario
In the second scenario, we have:
- 1 L of NaOH (1 M) mixed with 2 L of HCl (0.5 M).
- We need to find the temperature rise (ΔT2).

### Step 4: Calculate the total volume in the second scenario
The total volume (V2) in the second scenario is:
\[ V2 = 1 \, \text{L} + 2 \, \text{L} = 3 \, \text{L} \]

### Step 5: Establish the relationship between temperature change and volume
From thermodynamics, we know that the temperature change is inversely proportional to the volume when the number of moles of reactants remains constant. This can be expressed as:
\[ V1 \cdot \Delta T1 = V2 \cdot \Delta T2 \]

### Step 6: Substitute known values into the equation
We know:
- \( V1 = 2 \, \text{L} \)
- \( \Delta T1 = 10 \, \text{°C} \)
- \( V2 = 3 \, \text{L} \)

Substituting these values into the equation gives:
\[ 2 \cdot 10 = 3 \cdot \Delta T2 \]

### Step 7: Solve for ΔT2
Now, we can solve for ΔT2:
\[ 20 = 3 \cdot \Delta T2 \]
\[ \Delta T2 = \frac{20}{3} \]
\[ \Delta T2 \approx 6.67 \, \text{°C} \]

### Conclusion
The approximate rise in temperature when 1 L of NaOH (1 M) is mixed with 2 L of HCl (0.5 M) is **6.67°C**.

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