Question

10. In an effort to calculate the heat of neutralization of an acid, a student mixed 50.0 mL of 1.0 M H2SO4 with 100 mL of 1.0 M NaOH in a calorimeter and observes the temperature change. Write the balanced chemical equation for this reaction.

If the initial temperature of the acid and base was 20.1 °C and the temperature rose to 23.7 °C after mixing the two, what is the heat of neutralization for H2SO4?

Assume that the solutions have a density of 1.00 mg/mL and a specific heat of 4.184 J/g°C.

Answer #1

A balanced chemical equation for the neutralization reaction
between strong acid H_{2}SO_{4} and strong base
NaOH is:

2NaOH + H_{2}SO_{4} →
Na_{2}SO_{4} + 2H₂O

Moles of H_{2}SO_{4} = 0.050 L
H_{2}SO_{4} 1M (mol
H2SO_{4} 1L) = 0.050 mol
H_{2}SO_{4}

Volume of solution = (50.0 + 100) mL = 150.0 mL

Mass of solution = 150.0 mL soln 1.00g1mL soln = 150.0 g soln

ΔT = T_{2} – T_{1} = (23.7 – 20.1) °C = 3.6
°C

The heats involved are:

Heat from neutralization + Heat to warm solution + Heat to warm calorimeter = 0

i.e. q_{1}+q_{2}+q_{3} = 0

nΔH+mCΔT+CΔT = 0

0.050 mol ΔH + 150.0 g 4.184 J·g⁻¹°C⁻¹ 3.6 °C + 4.184 J·g⁻¹°C⁻¹ 3.6 °C = 0

0.050 mol ΔH + 2259.4 J + 15.1 J = 0

0.050 mol ΔH = - 2274.5 J

ΔH = − 2274.5 J 0.050mol = - 45490
J/mol = **- 45.5 kJ/mol**

Hence the heat of neutralization for H_{2}SO_{4}
is **- 45.5 kJ/mol**

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