Question

if 25 ml of 1 mol/L H2SO4 was added to 75 ml of 1 mol/L NaOH what is the expected temperature change of the solution?

Answer 1

We have reaction happening as:

H₂SO₄ + 2NaOH ------> Na₂SO₄ + 2H₂O

moles of sulfuric acid = molarity x volume in litres = 1mol/L x 0.025L = 0.025 mol

moles of NaOH = 1mol/L x 0.075L = 0.075 mol

1 mole sulfuric acid needs 2 mol sodium hydroxide to react completly

0.025 moles will need 2/1 x 0.025 = 0.050 mol NaOH to react completly

here we can see that sulfuric acid finishes first and hence it is limiting reactant and all the yield will be given by this only

ENTHALPY OF REACTION

[2ΔHf(H2O (ℓ)) + 1ΔHf(Na2SO4 (aq))] - [1ΔHf(H2SO4 (aq)) + 2ΔHf(NaOH (aq))]
[2(-285.83) + 1(-1389.47)] - [1(-909.27) + 2(-470.09)] = -111.68 kJ
-111.68 kJ     (exothermic)

for 0.025 mol H2SO4 = -111.68 kJ x 0.025 = -2.792 kJ = -2792 J

total volume = 25+75 = 100 mL

lets say its density = water density = 1g/mL

total mass = 100 mL x 1g/mL = 100g

heat absorbed by solution = mCp∆T = 2792J = 100g x 4.18 J/g.oC x ∆T

∆T = 6.68 ^oC