Calculate heat change

Question

In a constant‑pressure calorimeter, 58.7 mL of 0.930 M H₂SO₄ was added to 58.7 mL of 0.500 M NaOH.

The reaction caused the temperature of the solution to rise from 22.30 °C to 26.01 °C. If the solution has the same density and specific heat as water (1.00 g/mL and 4.184 J/(g⋅°C), respectively), what is ΔH for this reaction (per mole of H₂O produced) in kJ? Assume that the total volume is the sum of the individual volumes.

Answer 1

H₂SO₄ + 2NaOH ==> Na₂SO₄ + 2H₂O ... balanced equation

q = mC∆T

q = heat = ?

m = mass = 58.7 ml + 58.7 ml = 117.4 ml x 1 g/ml = 117.4 g

C = specific heat = 4.184 J/gº

∆T = change in temperature = 3.71º

q = (117.4 g)(4.184 J/gº)(3.71º) = -1822 J (negative sign because temp increased meaning exothermic rxn)

To find ∆H per mole H₂O, we need to find moles of H₂O produced:

moles H₂SO₄ present = 58.7 ml x 1 L/1000 ml x 0.930 mol/L = 0.0546 moles

moles NaOH present = 58.7 ml x 1 L /1000 ml x 0.50 mol/L = 0.0294

NaOH is limiting and dictates moles of H₂O formed.

moles H₂O formed = 0.0294 mol NaOH x 2 mol H₂O/2 mol NaOH = 0.0294 moles H₂O formed

∆H = -1822 J / 0.0294 mol H₂O = 61,973 J/mol = 62.0 kJ/mol (3 sig. figs.)