A 100.0 mL sample of 1.020 M HCl is mixed with a 50.0 mL sample of 2.040 M NaOH in a Styrofoam cup. If both solutions were initially at 24.53°C, and the enthalpy of the neutralization reaction is - 57 kJ/mole of H2O formed, what is the final temperature of the mixture? Assume that the solution has a density of 1.00 g/mL and a specific heat of 4.184 J/g°C, and that the Styrofoam cup has an insignificant heat capacity.

Question

Brycen Murillo

Chemistry

30 December, 22:38

A 100.0 mL sample of 1.020 M HCl is mixed with a 50.0 mL sample of 2.040 M NaOH in a Styrofoam cup. If both solutions were initially at 24.53°C, and the enthalpy of the neutralization reaction is - 57 kJ/mole of H2O formed, what is the final temperature of the mixture? Assume that the solution has a density of 1.00 g/mL and a specific heat of 4.184 J/g°C, and that the Styrofoam cup has an insignificant heat capacity.

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Tootsie · Answer 1

33.79°C

Explanation:

HCl and NaOH completely neutralize because they are strong acid and base.

In 100 mL of HCl we have 0.102 moles and in 50mL of NaOH we have 0.102 moles

As everything reacts, we have 0.102 moles of H2O.

The enthalpy of neutralization of 1 mole is - 57kJ, to 0,102 moles is - 5.814 kJ. The reaction is exothermic.

Now, we have a 150 mL sample, as the density is 1.00 g/ml, we have 150 g of solution, with the specific heat of 4.184 J/g°C.

With the equation Q = mcΔT, as Q is the heat (5,817 J), m is the mass (150 g) c is the specific heat (4.184 J/g°C) and ΔT is the temperature variation.

5,814 = 150 x 4.184 x ΔT

ΔT = 9.26 °C

T = Ti + ΔT = 24.53 + 9.26 = 33.79 °C