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.

Question

Natalya Mckee

Chemistry

10 September, 17:35

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.

+3

Answers (1)

Know the Answer?

Hudson Shea · Answer 1

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

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