Question

In a constant-pressure calorimeter, 75.0 mL of 0.810 M H2SO4 was added to 75.0 mL of 0.480 M NaOH. The reaction caused the temperature of the solution to rise from 24.47 °C to 27.74 °C. If the solution has the same density and specific heat as water (1.00 g/mL and 4.184 J/g·K, respectively), what is ΔH for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes.

Answer 1

Balanced equation:
2 NaOH + H₂SO₄ ===> Na₂SO₄ + 2 H₂O
Reaction type: double replacement

Moles of H2SO4 reacted = 75 x 0.81 /1000 = 0.06075 Moles

Moles of NaOH reacted = 75 x 0.81 /1000 = 0.036 Moles

NaOH is Limiting reagent

Moles of water produced = 0.036 Moles

Let us find the heat for this mole

Q = mc∆T

Q = heat energy (Joules, J), m = mass of a substance (kg)

c = specific heat (units J/kg∙K), ∆ is a symbol meaning "the change in"

∆T = change in temperature (Kelvins, K)

Q = 150 x 4.184 x (27.74 - 24.47) = 2052.252 Joules

ΔH for this reaction = 2052.252 / 0.036 x 1000 = 57.007 Kilo Joules