Question

A common laboratory reaction is the neutralization of an acid with a base. When 49.2 mL of 0.500 M HCl at 25.0°C is added to 59.2 mL of 0.500 M NaOH at 25.0°C in a coffee cup calorimeter (with a negligible heat capacity), the temperature of the mixture rises to 28.2°C. What is the heat of reaction per mole of NaCl (in kJ/mol)? Assume the mixture has a specific heat capacity of 4.18 J/(g·K) and that the densities of the reactant solutions are both 1.07 g/mL.

Answer 1

To find the heat of reaction per mole of NaCl for the neutralization of HCl with NaOH, the heat absorbed by the solution is calculated and then divided by the number of moles of NaCl produced, resulting in approximately 31.37 kJ/mol.

Step-by-step explanation:

The student is asked to determine the heat of reaction per mole of NaCl for the neutralization of HCl with NaOH in a coffee cup calorimeter. To find this, we first need to calculate the total heat released using the formula q = mcΔT, where m is the mass of the solution, c is the specific heat capacity, and ΔT is the change in temperature. Given that the densities of both solutions are 1.07 g/mL, we can calculate the mass by multiplying the volume of the solution in milliliters by its density.

We have (49.2 mL + 59.2 mL) × 1.07 g/mL as the total mass of the solution which comes out to be 115.68 g. The temperature change ΔT is 28.2°C - 25.0°C = 3.2°C. The specific heat capacity c is given as 4.18 J/(g·K). Thus, the heat absorbed by the solution is q = 115.68 g × 4.18 J/(g·K) × 3.2 K, which equals 1543.3536 J or 1.543 kJ.

Since the reaction involves 0.0492 mol of NaCl (limited by the volume and molarity of HCl), we divide the total heat by the moles to find the heat of reaction per mole of NaCl: 1.543 kJ ÷ 0.0492 mol, which gives us approximately 31.37 kJ/mol.

Answer 2

The heat of reaction per mole of NaCl is Q = 6.204 kJ/mol.

Step-by-step explanation:

HCl + NaOH → NaCl + H₂O

V: 49.2 mL 59.2 mL

C: 0.500 M 0.500 M

n = CV

n= 0.025 mol 0.03 mol → 0.025 mol

In this reaction, 0.025 mol of NaCl is produced.

Q = mcΔT

m is the total mass od the mixture.

density = mass/volume ∴ mass = density x volume

m(HCl) = 1.07x49.2 = 52.64g

m(NaOH) = 1.07x59.2 = 63.4g

m = m(HCl) +m(NaOH) = 115.98g

Q = 115.98 g x 4.18 J/gK x 3.2K

Q = 1.551 kJ

This value is for 0.025 mol of NaCl. Therefore, for 1 mol of NaCl we need four times this value.

Q = 4 x 1.551

Q = 6.204 kJ/mol