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in a coffee cup calorimeter, 100.0 ml of 1.0 m naoh and 100.0 ml of 1.0 m hcl are mixed. both solutions were orignially at 24.68c. after the reaction, the final temperature is 31.38c. assuming that all the solutions have a density of 1.0 g/cm3 and a specific heat capacity of 4.18 j 8c21 g21, calculate the enthalpy change for the neutralization of hcl by naoh. assume that no heat is lost to the surroundings or to the calorimeter

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Final answer:

To find the enthalpy change for the neutralization reaction of HCl by NaOH, calculate the heat released using the mass, specific heat capacity, and temperature change, and then divide by the number of moles reacting.

Step-by-step explanation:

To calculate the enthalpy change for the neutralization of HCl by NaOH in a coffee cup calorimeter, we use the information provided about the initial and final temperatures, the volume and molarity of the solutions, and the assumption that the density and specific heat capacity are the same as those of water. Since the volumes of NaOH and HCl solutions are both 100.0 mL, their combined volume is 200.0 mL. Given that the density is 1.0 g/cm³, the mass of the combined solutions is 200.0 g. The specific heat capacity is 4.18 J/g°C. The heat (q) absorbed by the solution is calculated using the formula: q = m × c × ΔT, where m is mass, c is specific heat capacity, and ΔT is the change in temperature.

The temperature change (ΔT) is the final temperature minus the initial temperature, which is 31.38°C - 24.68°C = 6.7°C. Plugging the values into the formula we get q = 200.0 g × 4.18 J/g°C × 6.7°C, which calculates the heat released during the reaction.

Finally, to find the enthalpy change per mole (ΔH), we divide the heat produced by the number of moles reacting. Since we started with 1.0 M solutions of NaOH and HCl, and each had a volume of 100 mL, we have 0.1 mol of each reactant. Dividing the total heat by 0.1 mol will give us ΔH in kJ/mol.

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