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. What is the heat of reaction per mole of acid? Assume the mixture has a specific heat capacity of 4.18 J/g·°C and that the densities of the reactant solutions are both 1.00 g/mL.
A) 670 J
B) 1300 J
C) 27 kJ
D) 54 kJ
E) 100 kJ

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. What is the heat of reaction per mole of acid? Assume the mixture has a specific heat capacity of 4.18 J/g·°C and that the densities of the reactant solutions are both 1.00 g/mL.
A) 670 J
B) 1300 J
C) 27 kJ
D) 54 kJ
E) > 100 kJ

Quizplus · Accepted Answer

The heat of reaction can be calculated using the equation:

q = mcΔT

where q is the heat absorbed or released by the reaction, m is the mass of the mixture, c is the specific heat capacity of the mixture, and ΔT is the change in temperature.

First, we need to calculate the mass of the mixture. Since we have equal volumes of the two solutions with the same density, we can assume that they have the same mass:

mass = volume x density = 50.0 mL x 1.00 g/mL + 50.0 mL x 1.00 g/mL = 100 g

Next, we need to calculate the change in temperature:

ΔT = Tf - Ti = 28.2°C - 25.0°C = 3.2°C

Now we can substitute these values into the equation:

q = (100 g) x (4.18 J/g·°C) x (3.2°C) = 1337.6 J

However, this is the heat absorbed/released by 100 g of the mixture, and we want to know the heat per mole of acid. Since we have equal concentrations of HCl and NaOH, the limiting reactant is HCl, and one mole of HCl reacts with one mole of NaOH. Therefore, the number of moles of HCl in 50.0 mL of 0.500 M solution is:

moles HCl = (0.500 mol/L) x (0.0500 L) = 0.0250 mol

So the heat of reaction per mole of acid is:

(1337.6 J) / (0.0250 mol) = 53,504 J/mol ≈ 54 kJ/mol

Therefore, the answer is D) 54 kJ.

A Common Laboratory Reaction Is the Neutralization of an Acid