change in enthalpy

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. Note: do not include units in your response.

Answer 1

q = mC∆T

At constant pressure ∆H = q

q = ∆H = heat or change in enthalpy = ?

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 = 26.01 - 22.30 = 3.71º

∆H = (117.4 g)(4.184 J/gº)(3.71º)

∆H = 1822 J this is the change in enthalpy for the reaction, but to express it on a per mole of H2O basis, we need to look at the balanced equation and find out how many moles of water were produced.

H2SO4 + 2NaOH ==> Na2SO4 + 2H2O

moles H2SO4 = 0.0587 L x 0.930 mol/L = 0.0797 moles

moles NaOH = 0.0587 L x 0.500 mol/L = 0.04285 moles and this is LIMITING

moles of H2O produced from 0.04285 moles NaOH = 0.04285 (1:1 mol ratio in balanced equation)

∆H in kJ/mol H2O = 1822 J / 0.04285 moles x 1 kJ/1000 J = 42.5 kJ/mole (to 3 sig. figs.)