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Calculating Enthalpy Change?
In a coffee cup calorimeter, 104.0 mL of 1.9 M NaOH and 104.0 mL of 1.9 M HCl are mixed. Both solutions were originally at 22.3°C. After the reaction, the final temperature is 35.0°C. Assuming that all the solutions have a density of 1.0 g/cm3 and a specific heat capacity of 4.18 J·°C−1·g−1,
calculate the enthalpy change (in kJ/mol) for the neutralization of HCl by NaOH. Assume that no heat is lost to the surroundings or to the calorimeter.
1 Answer
- Roger the MoleLv 72 weeks ago
(4.18 J/g·°C) x (104.0 mL + 104.0 mL) x (1.0 g/mL) x (35.0 - 22.3)°C =
11041.888 J produced
(11041.888 J) / (0.1040 L x 1.9 mol/L) = 55880 J / mol = - 55.9 kJ/mol
[The minus sign did not come from the calculations. It is the convention for exothermic processes.]
