Enthalpy Change

Question

What is the enthalpy change (in kJ) for the reaction of 350.0 mL of a 0.480 M NaOH

solution with 275 mL of a 0.360 M H2SO4 solution? (7)

2 NaOH(aq) + H2SO4(aq) ® Na2SO4(aq) + 2 H2O(l) 

Hrxn = –112 kJ

enthalpy change (kJ): _____________

Answer 1

2NaOH(aq) + H₂SO₄(aq) ==> Na₂SO₄(aq) + 2H₂O(l)

moles NaOH present = 0.3500 L x 0.480 mol/L = 0.168 moles NaOH

moles H₂SO₄ present = 0.275 L x 0.360 mol/L = 0.0990 moles H₂SO₄

Limiting reactant = NaOH since the mole ratio in the balanced eq. shows 2 moles NaOH for each mol H₂SO₄

The enthalpy of reaction is generally given in units of energy per amount of substance, e.g. kJ/mole. You provided a value of -112 kJ without units of amount of substance. We will take that as -112 kJ/2 moles NaOH as in the balanced equation. Thus,

-112 kJ/2 moles x 0.168 moles = 9.41 kJ