Question

A common laboratory reaction is the neutralization of an acid with a base. When 50.0 mL of 0.500 M HCl at 25.0°C is added to 50.0 mL of 0.500 M NaOH at 25.0°C in a coffee cup calorimeter, the temperature of the mixture rises to 28.2°C. Assume the mixture has a specific heat capacity of 4.18 J/(g.K) and the densities of all solutions are 1.00 g/mL

A.) Calculate the qrxn
B.) Calculate deltaH rxn per mole of acid
C.) When this reaction is repeated with 50.0 mL of 2.0 M HCI at 25.0°C, the temperature of the mixture also increases to 28.2°C. Explain

Answer 1

a) qrxn = 1,337.6 J

b) ΔHrxn /mol = -5.35 kJ/mol

c) See explanation below

Step-by-step explanation:

a) q rxn = mcΔT where m is the mass of water, c its specific heat capacity , and ΔT is the change in temperature.

m H₂O = (50.0 mL + 50.0 mL) 1.00 g/mL= 100 g

q rxn = 100 g x 4.18 J/gK x (28.2 -25.0) K = 1,337.6 J

b) ΔHrxn = - qrxn

mol HCl =( 50mL/1000mL/L )x 0.500 mol/L = 0.025 mol

ΔHrxn / mol = -1337.6 J/0.025 mol = -53,504.0 J = -5.35 kJ/mol

c) We just calculated the heat of neutralization per mol HCl, but thats not the whole story.

We have to ensure that if we increase the quantity of HCl ( 0.100 mol) we also increase the amount NaOH in the same proportion, i.e. 0.100, if not only 0.025 mol HCl will react liberating the same amount of heat calculated above.