The enthalpy change when a strong acid is neutralized by strong base is -56.1 kJ/mol. If 135 mL of 0.450 M HI at 23.15°C is mixed with 145 mL of 0.500 M NaOH, also at 23.15°C, what will be the maximum temperature reached by the resulting solution? (Assume that there is no heat loss to the container, that the specific heat of the final solution is 4.18 J/g·°C, and that the density of the final solution is that of water.)

Question

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The correct answer is A because the enthalpy change of the reaction is negative, which means that heat is released. The amount of heat released can be calculated using the equation Q = -ΔH * n, where Q is the heat released, ΔH is the enthalpy change, and n is the number of moles of reactants. In this case, the number of moles of HI is 0.135 L * 0.450 M = 0.0608 mol, and the number of moles of NaOH is 0.145 L * 0.500 M = 0.0725 mol. The limiting reactant is HI, so the number of moles of heat released is 0.0608 mol * -56.1 kJ/mol = -3.42 kJ. The mass of the final solution is 135 mL + 145 mL = 280 mL = 0.280 kg. The temperature change can be calculated using the equation Q = mcΔT, where m is the mass of the solution, c is the specific heat of the solution, and ΔT is the temperature change. In this case, the temperature change is -3.42 kJ / (0.280 kg * 4.18 J/g·°C) = 26.06°C. The maximum temperature reached by the resulting solution is 23.15°C + 26.06°C = 49.21°C.

The Enthalpy Change When a Strong Acid Is Neutralized by Strong