To find the final temperature of the reaction when 25 mL of 1.0 M HCl is combined with 35.0 mL of 0.5 M NaOH, we can follow these steps:
### Step 1: Calculate the moles of HCl and NaOH
1. **Moles of HCl**:
\[
\text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 1.0 \, \text{mol/L} \times 0.025 \, \text{L} = 0.025 \, \text{mol}
\]
2. **Moles of NaOH**:
\[
\text{Moles of NaOH} = \text{Concentration} \times \text{Volume} = 0.5 \, \text{mol/L} \times 0.035 \, \text{L} = 0.0175 \, \text{mol}
\]
### Step 2: Determine the limiting reactant
- The reaction between HCl and NaOH is a 1:1 reaction. Since we have 0.025 mol of HCl and 0.0175 mol of NaOH, NaOH is the limiting reactant.
### Step 3: Calculate the heat released (Q) during the reaction
- The standard enthalpy change for the reaction is given as -56 kJ/mol. Since NaOH is the limiting reactant:
\[
Q = \text{moles of NaOH} \times \Delta H = 0.0175 \, \text{mol} \times (-56 \, \text{kJ/mol}) = -0.98 \, \text{kJ} = -980 \, \text{J}
\]
### Step 4: Calculate the total mass of the solution
- The total volume of the solution is:
\[
\text{Total Volume} = 25 \, \text{mL} + 35 \, \text{mL} = 60 \, \text{mL}
\]
- Since the density of the solution is 1 g/mL, the mass of the solution is:
\[
\text{Mass} = \text{Density} \times \text{Volume} = 1 \, \text{g/mL} \times 60 \, \text{mL} = 60 \, \text{g}
\]
### Step 5: Calculate the temperature change (ΔT)
- Using the formula \( Q = m \cdot c \cdot \Delta T \):
\[
-980 \, \text{J} = 60 \, \text{g} \times 4184 \, \text{J/(g°C)} \times (T_f - 25°C)
\]
\[
-980 = 250,440 \times (T_f - 25)
\]
### Step 6: Solve for the final temperature (Tf)
1. Rearranging the equation:
\[
T_f - 25 = \frac{-980}{250440}
\]
\[
T_f - 25 = -0.00391
\]
\[
T_f = 25 - 0.00391 = 24.9961°C
\]
### Step 7: Final Temperature
- The final temperature of the solution after the reaction is approximately:
\[
T_f \approx 25.00°C
\]
### Summary
The final temperature of the reaction is approximately **25.00°C**.
