Acommon 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

a) qrxn = 1,337.6 J

b)  ΔHrxn /mol = -5.35 kJ/mol

c) See explanation below

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.