Question

A coffee cup calorimeter having a heat capacity of 451 J/ o C was used to measure the heat evolved when 0.0300 mol of NaOH(s) was added to 1000 mL of 0.0300 M HNO 3 initially at 23.000 o C. The temperature of the water rose to 23.639 o C. Calculate H (in kJ/mol NaNO 3 ) for this reaction. Assume the specific heat of the final solution is 4.18 J/g o C; the density of each solution is 1.00 g/mL; and the addition of solid does not appreciably affect the volume of the solution. HNO 3 (aq) NaOH(s) NaNO 3 (aq) H 2 O(l)

Answer 1

To calculate the enthalpy change (ΔH) for the formation of NaNO3 from the reaction of NaOH and HNO3 in a calorimeter, first calculate the total heat evolved using the specific heat of the solution and the calorimeter heat capacity, then divide by the moles of NaOH reacted to find ΔH per mole in kJ/mol.

To calculate the enthalpy change (ΔH) for the reaction between sodium hydroxide (NaOH) and nitric acid (HNO3) forming sodium nitrate (NaNO3) and water (H2O), we can use the data from the coffee cup calorimeter experiment. Fist, we determine the total heat evolved Q using the specific heat capacity (c), the mass of the solution (m), and the change in temperature (ΔT). Applying Q = mcΔT considering the calorimeter’s heat capacity, we have:

Q = (specific heat of solution) × (mass of the solution) × (change in temperature) + (heat capacity of calorimeter) × (change in temperature)

Q = (4.18 J/g°C) × (1000 g) × (23.639 °C – 23.000 °C) + (451 J/°C) × (23.639 °C – 23.000 °C)

Q = (4.18 J/g°C) × (1000 g) × (0.639 °C) + (451 J/°C) × (0.639 °C)

After calculating Q, which is the total heat evolved for the reaction of 0.0300 mol NaOH, we can find the enthalpy change per mole by ΔH = Q / moles of NaOH, thus:

ΔH = Q / 0.0300 mol

Convert Q into kilojoules by dividing it by 1000, then calculate ΔH. The enthalpy change will have a negative sign, indicating an exothermic reaction. ΔH is reported in kJ/mol NaNO3.