Question

At constant pressure, 50.0 mL of 0.100 M KOH and 45.0 mL of 0.100 M HNO3 are mixed in a styrofoam cup.The two solutions are initially at 25.00 °C and the final temperature is 25.65 °C. Calculate the ΔHneutralization (kJ/mol) in term of moles of HNO3 since it’s the limiting reagent. Assume a specific heat capacity of 4.184 J/(°C•g) and that the density of the solution is the same as water (0.997 g/mL).

Answer 1

≈ -57.2 kJ/mol

Step-by-step explanation:

Total volume of the solution = (50.0 + 45.0) mL = 95.0 mL.

Density of the solution = 0.997 g/mL.

Mass of the solution = (volume of the solution)*(density of the solution)

= (95.0 mL)*(0.997 g/mL)

= 94.715 g

Heat gained by the solution, qsoln = (mass of the solution)*(specific heat capacity)*(change in temperature)

= (94.715 g)*(4.184 J/ºC.g)*(25.65 – 25.00)ºC

= 257.5869 JAccording to the principle of thermochemistry,

qsoln + qrxn = 0

where qrxn denotes the heat change during the neutralization reaction.

Therefore,

(257.5869 J) + qrxn = 0

======> qrxn = -257.5869 J

HNO3 is the limiting reactant.

Moles of limiting reactant = (volume of the limiting reactant in L)*(molarity of limiting reactant)

= (45.0 mL)*(0.100 M)

= (45.0 mL)*(1 L)/(1000 mL)*(0.100 M)

= 0.0045 mole.

ΔHneutralization = qrxn/(moles of HNO3)

= (-257.5869 J)/(0.0045 mol)

= -57241.5 J/mol

= (-57241.5 J/mol)*(1 kJ)/(1000 J)

= -57.2415 kJ/mol

≈ -57.2 kJ/mol