To solve the problem step by step, let's analyze the situation:
### Step 1: Calculate the moles of `H2SO4` added
We have 500 mL of 0.1 M `H2SO4`. To find the number of moles, we use the formula:
\[
\text{Moles of } H2SO4 = \text{Volume (L)} \times \text{Molarity (mol/L)}
\]
\[
\text{Moles of } H2SO4 = 0.5 \, \text{L} \times 0.1 \, \text{mol/L} = 0.05 \, \text{mol}
\]
### Step 2: Calculate the moles of `NaOH` present
We have 1 L of 0.1 M `NaOH`. Similarly, we calculate the moles:
\[
\text{Moles of } NaOH = 1 \, \text{L} \times 0.1 \, \text{mol/L} = 0.1 \, \text{mol}
\]
### Step 3: Determine the neutralization reaction
The reaction between `H2SO4` and `NaOH` can be represented as:
\[
H2SO4 + 2 NaOH \rightarrow Na2SO4 + 2 H2O
\]
From the reaction, we see that 1 mole of `H2SO4` reacts with 2 moles of `NaOH`.
### Step 4: Identify the limiting reactant
We have 0.05 moles of `H2SO4` and 0.1 moles of `NaOH`. According to the stoichiometry:
- To completely neutralize 0.05 moles of `H2SO4`, we need \(0.05 \times 2 = 0.1\) moles of `NaOH`.
Since we have exactly 0.1 moles of `NaOH`, both reactants will completely neutralize each other.
### Step 5: Calculate the heat evolved during the first neutralization
The heat evolved during the neutralization is given as `x` calories. Since the reaction goes to completion, we can say that all available acid and base have reacted.
### Step 6: Add more `H2SO4` to the solution
Now, we add another 500 mL of 0.1 M `H2SO4`, which again gives us:
\[
\text{Moles of additional } H2SO4 = 0.5 \, \text{L} \times 0.1 \, \text{mol/L} = 0.05 \, \text{mol}
\]
### Step 7: Analyze the new situation
After the first neutralization, we have no `NaOH` left (it was completely neutralized). Now, we are adding 0.05 moles of `H2SO4` to a solution that has no base left to react with. Therefore, there will be no further neutralization reaction.
### Step 8: Conclusion on heat evolved
Since no further neutralization occurs, the heat evolved from adding the second 500 mL of `H2SO4` will be 0 calories.
### Final Answer
The heat evolved after adding the second 500 mL of `H2SO4` will be **0 calories**.
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