Question

In a constant‑pressure calorimeter, 60.0 mL of 0.920 M H 2 SO 4 was added to 60.0 mL of 0.360 M NaOH . The reaction caused the temperature of the solution to rise from 22.43 ∘ C to 24.88 ∘ 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 Δ H for this reaction (per mole of H 2 O produced)? Assume that the total volume is the sum of the individual volumes.

Answer 1

First to say, mL x M = mL x mol / L = (mL / L) x mol = 10^-3 mol

Since NaOH used is much lower than the H₂SO₄, so NaOH is the terminal reagent and it will decide the amount of water produced. Since one mole of NaOH furnishes one mole of OH^- , so water produced will be the no. of moles of NaOH.

60.0 mL 0.360 M NaOH = (60.0 x 0.360 x 10^-3) mol = 0.0216 mol. So, this much moles of H₂O is produced.

Total volume of the mixture is (60.0 + 60.0) mL = 120.0 mL. Using density as 1.00 g/mL, mass of the solution is m = 120.0 g.

Specific heat of the solution = s = 4.184 J / (g. ^oC)

change in temperature = = (24.88 - 22.43) ^oC = 2.45 ^oC.

So, heat released due to the reaction is = ms = [ 120.0 g x 4.184 J / (g. ^oC) x 2.45 ^oC ] = 1230.096 J.

If no heat is lost to the surroundings, then this is produced by the heat released by the neutralization reaction occured. Now, if heat released due to the reaction for producing one mole of water is , then total heat released is = { 0.0216 mol x }

Now,

Since the reaction is exothermic, so the quatity is negative, thus,