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.

Answer 1

This sounds like a molar enthalpy problem. You can use the equation:  ΔH = q = mcΔT. In this specific case, H=q because pressure is constant. You would first need to recognize you were given mL and molarity which can be used to find the grams of each substance.

Given is d = 1.00 g/mL and the solution is 117.4 mL (58.7 mL + 58.7 mL), m= v*d --> 117.4 grams of solution. ΔH of water = 4.184 J/g°C

Calculate ΔT = Tf - Ti

Plug knowns into ΔH = q = mcΔT to get heat gained by water.

But, question is looking for per mole of H2O... therefore, take the ΔH=q calculated and divide by moles of H2O which can be found via balancing the equation/reaction/stoichiometry (__H2SO4 + __NaOH --> __H2O + __Na2SO4) .

Hint: to find moles of each substance, multiply molarity by volume of substance. (M= moles/liter)

Hope this helps :)