Question

# 10. In an effort to calculate the heat of neutralization of an acid, a student mixed...

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.

A balanced chemical equation for the neutralization reaction between strong acid H2SO4 and strong base NaOH is:

2NaOH + H2SO4   →   Na2SO4 + 2H₂O

Moles of H2SO4 = 0.050 L H2SO4 1M (mol H2SO4 1L) = 0.050 mol H2SO4

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 = T2 – T1 = (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. q1+q2+q3 = 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 H2SO4 is - 45.5 kJ/mol

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