The enthalpy change when a strong acid is neutralized by strong base is -56.1 kJ/mol. If 12.0 mL of 6.00 M HBr at 21.30°C is mixed with 300.mL of 0.250 M NaOH, also at 21.30°C, what will 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.)

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Quizplus · Accepted Answer

The heat released by the neutralization reaction will raise the temperature of the solution. The amount of heat released can be calculated from the enthalpy change and the number of moles of acid and base that react. The temperature change can then be calculated using the specific heat and density of the solution.The number of moles of HBr is:$$12.0 	ext{ mL} 	imes \frac{1 	ext{ L}}{1000 	ext{ mL}} 	imes 6.00 	ext{ M} = 0.0720 	ext{ mol}$$The number of moles of NaOH is:$$300. 	ext{ mL} 	imes \frac{1 	ext{ L}}{1000 	ext{ mL}} 	imes 0.250 	ext{ M} = 0.0750 	ext{ mol}$$The limiting reactant is HBr, so the number of moles of NaOH that react is also 0.0720 mol.The heat released by the reaction is:$$0.0720 	ext{ mol} 	imes (-56.1 	ext{ kJ/mol}) = -4.03 	ext{ kJ}$$The mass of the final solution is:$$(12.0 	ext{ mL} + 300. 	ext{ mL}) 	imes \frac{1 	ext{ g}}{1 	ext{ mL}} = 312 	ext{ g}$$The temperature change is:$$\frac{-4.03 	ext{ kJ}}{312 	ext{ g} 	imes 4.18 	ext{ J/g} \cdot 	ext{°C}} = -3.03 	ext{ °C}$$The maximum temperature reached by the solution is:$$21.30 	ext{ °C} - 3.03 	ext{ °C} = 18.27 	ext{ °C}$$The closest answer choice is D) 24.40°C.

The Enthalpy Change When a Strong Acid Is Neutralized by Strong