n a coffee cup calorimeter, 100.0 mL of 1.0M NaOH and 100.0 mL of 1.0M HCl are mixed. Both solutions were originally at 24.6 degrees Celsius. After the reaction, the final temperature is 31.3 degrees Celsius. Assuming that all the solution have a density of 1.0g/cm3 and a specific heat capacity of 4.18 J/oCg, calculate the enthalpy change for the neutralization of HCl by NaOH. Assume that no heat is lost to the surroundings or to the calorimeter.

### Answers

56.0kJ/mol

Explanation:

The reaction of NaOH with HCl is:

NaOH + HCl → H₂O + NaCl + ΔH

Where ΔH is the heat of reaction that is released per mole of reactants,

The moles that reacts are:

100mL = 0.1L * (1mol / L) = 0.1 moles reacts

To find the heat released in the coffee cup calorimeter, you use the equation:

Q = m×ΔT×C

Where Q is heat released,

m is mass of the solution

ΔT is change in temperature (Final temperature - Initial temperature)

C is specific heat of the solution (4.18J/g°C)

Mass of the solution is:

100mL + 100mL = 200mL

Density of the solution is 1.0g/mL. The mass is 200g

Change in temperature is 31.3°C - 24.6°C = 6.7°C

Replacing:

Q = m×ΔT×C

Q = 200g×6.7°C×4.18J/g°C

Q = 5601.2J

This is the heat released per 0.1mol. The heat released per mole (Enthalpy change for the neutralization of HCl by NaOH is:

5601.2J / 0.1 moles = 56012J / mol =

56.0kJ/mol

Explanation:

mass of the solution = volume x density = 200 x 1 = 200 gm

heat absorbed = m x s x Δ t , s is specific heat , Δt is rise in temperature

= 200 x 4.18 x ( 31.3 - 24.6 )

= 5601 J .

This is the enthalpy change required.