Question

In a constant-pressure calorimeter, 60.0 mL of 0.760 M H2SO4 was added to 60.0 mL of 0.500 M NaOH. The reaction caused the temperature of the solution to rise from 23.05 °C to 26.46 °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.

I am stuck on this problem!!

Answer 1

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

Moles of H₂SO₄ reacting = molarity*volume of solution in litres = 0.76*0.06 = 0.0456

moles of NaOH reacting = molarity*volume of solution in litres = 0.5*0.06 = 0.03

Clearly, from the above balanced reaction, H₂SO₄ and NaOH reats in the molar ratio of 1:2

Clearly, NaOH is the limiting reagent

Now, moles of H₂SO₄ reacting = 0.5*moles of NaOH reacting = 0.015

Molar mass of H₂SO₄ = 98 g/mole and molar mass of NaOH = 40 g/mole

Thus, mass of solution reacting = 0.015*98 + 0.03*40 = 2.67 g

Thus, heat generated in the reaction, delta H = mass*specific heat*change in temperature = 2.67*4.184*3.41 = 38.09 J