Final answer:
The enthalpy change for the reaction of 0.430 M sulfuric acid with 0.309 M potassium hydroxide was calculated using moles and the heat of neutralization; however, the calculated value does not match the options provided, suggesting a need to verify the data used.
Step-by-step explanation:
To calculate the enthalpy change when 0.430 M sulfuric acid reacts with 0.309 M potassium hydroxide, we must first determine the limiting reactant. We can calculate the moles of each reactant:
- For H2SO4: 0.0494 L × 0.430 mol/L = 0.0212 mol
- For KOH: 0.0233 L × 0.309 mol/L = 0.0072 mol
The balanced reaction is:
H2SO4 + 2KOH → K2SO4 + 2H2O
KOH is the limiting reactant because it requires twice as many moles of KOH as H2SO4 for the reaction to occur. Therefore, only 0.0072 mol of H2SO4 will react, and the molar ratio is 1:1 for the reaction that occurs.
Now, assuming the heat of neutralization for a strong acid with a strong base is similar to the provided example, -58 kJ/mol, and that the heat release is proportional to the amount of reactant, we can calculate the enthalpy change (ΔH):
ΔH = -58 kJ/mol × 0.0072 mol = -0.4176 kJ
Since none of the options closely match our calculated value, we should check that the provided information and our calculations are appropriate for the given question.