Question

100 mL of 0.500 M HCl is mixed with 100 mL of 0.600 M NaOH in a constant pressure calorimeter. The initial temperature of the solutions is 22.50 C and the final temperature of the mixed solution is 25.86 C. Calculate ?_rH (in units of JK/mol-rxn) for the reaction

NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)

The density of the resulting solution is 100 g/mL, and its specific heat is 4.184 J/(g C).

Thank you in advance!

Answer 1

Ans. #1. Step 1: Determine the limiting reactant:

Moles of HCl = Molarity x Vol of soln. in liters

                                    = 0.500 M x 0.100 L = 0.050 mol

# Moles of NaOH = 0.600 M x 0.100 L = 0.060 mol

# Balanced reaction:             HCl(aq) + NaOH(aq) ----> NaCl(aq) + H₂O(l)

Or:                               H⁺(aq) + OH^-(aq) --------> H₂O(aq)

Following stoichiometry, 1 mol H⁺ (from HCl) neutralizes 1 mol OH^- (from NaOH) to form 1 mol H₂O.

So, Theoretical molar ratio of reactant: HCl : NaOH = 1 : 1

# Experimental molar ratio of reactant: HCl : NaOH = 0.05 : 0.06 = 0.8 : 1

# Comparing the theoretical and experimental molar ratios of reactants, the experimental moles of HCl is less than its theoretical value of 1 while that of NaOH is kept constant at 1 mol.

So, HCl is the limiting reactant.

# The formation of product (H₂O) follows the stoichiometry of limiting reactant.

So,

            Moles of water of neutralization formed = 0.050 mol

# Step 2: Assuming density of 1.00 g/mL            –

the mass of mixture = 100.0 g + 100.0 g = 200.0 g

            dT = Final temperature – Initial temperature

Step 2: Heat lost by solution is given by Equation 1 Where, s specific heat of water4.184 J g1 oc Putting the values in equation 1 Or, q-200.00 g x 4.184 Jg1°c Hence, q X 3.36 °c 2811.65 J

# Step 3: The total heat gain by solution must be equal to the total heat released during neutralization.

So,

            Formation of 0.050 mol H₂O releases 2811.65 J energy.

Now,

            dH = Energy released / Moles of H₂O formed

                        = -2811.65 J / 0.050 mol

                        = - 56233.0 J/mol

                        = -56.233 kJ/mol

Note: The –ve sign indicates release of heat.

Answer 2

A quantity of 400 mL of 0.600 M HCl was mixed with 400 mL of 0.600 M NaOH in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the HCl and NaOH solutions was the same, 27.50 °C, and the final temperature of the mixed solution was 32.45 °C. Calculate the heat change for the neutralization reaction on a molar bass basis NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l) Assume that the densities and specific heats of solutions are the same as for water (1.00 g/mL and 4.184 J/g∙ °C, respectively)