Question

A student adds 75.0 mL of a 1.80MHCl to 75.0 mL of 1.80MNaOH inside a calorimeter (C= 77.0 J/° C ) and immediately places the lid on top. The solution temperature rises from 20.5°C to 24.7°C. The specific heat of water in 4.184 J/g° C and the density of water is 1.00 g/mL. Assume the density of the acid and base solutions are also 1.00 g/mL.

Calculate the heat of reaction and calculate the molar enthalpy for the reaction.

Answer 1

The heat of reaction for the neutralization of HCl and NaOH in a calorimeter is -2635.92 J, and the molar enthalpy for the reaction is -19.525 kJ/mol, indicating an exothermic reaction.

Step-by-step explanation:

To calculate the heat of reaction and the molar enthalpy for the reaction of HCl and NaOH in a calorimeter, we use the following relationships and data given:

The heat absorbed by the solution (q) can be calculated using the formula:

q = mass × Cs × ΔT

q = 150.0 g × 4.184 J/g°C × 4.2°C

q = 2635.92 J

Since the reaction is exothermic, the heat of reaction released by the reaction is -2635.92 J.

To find the molar enthalpy, we need to determine the number of moles of HCl and NaOH that reacted. Since both are 1.80 M and the volume is 75.0 mL or 0.075 L for each, the moles of each are:

moles HCl = moles NaOH = 1.80 M × 0.075 L = 0.135 mol

The molar enthalpy (ΔH) is calculated by dividing the heat by the number of moles:

ΔH = q / moles

ΔH = -2635.92 J / 0.135 mol

ΔH = -19525.33 J/mol or -19.525 kJ/mol (since 1 kJ = 1000 J)

The negative sign indicates that the reaction is exothermic, releasing heat to the surroundings.