question is below in the description

Question

In a constant-pressure calorimeter, 55.0 mL of 0.310 M Ba(OH)2 was added to 55.0 mL of 0.620 M HCl. The reaction caused the temperature of the solution to rise from 24.15 °C to 28.37 °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

Answer 1

Hi Sheridan!

Questions like this one usually involve multiple calculations that you combine at the end for a final answer. I like to start by imagining what my answer will be. In this case the units of the answer will be in Joules/mol H₂O. That means I'll need to calculate the heat (J) and then mol H₂O produced.

Heat:

The temperature change was 24.37 to 24.15 ^oC, which is a delta T of 4.22^oC or 4.22K. Since the volumes are additive the total volume of the final solution is 110.0 mL. Multiply by the density of 1.00g/mL and we have 110.0g H₂O.

Heat = (4.184 J/gK)(110.0g H₂O)(4.22K) = 1,942.21 J (Keeping extra Sig Figs until the end)

Mol H₂O:

First write the balanced equation:

2HCl + Ba(OH)₂ ----> BaCl₂ + 2H₂O

mol H₂O = (0.0550L)(0.310 M Ba(OH)₂)(2 H₂O/1 Ba(OH)₂) = 0.0341 mol H₂O

mol H₂O = (0.0550L)(0.620 M HCl)(1 H₂O/1 HCl) = 0.0341 mol H₂O

The reactants are perfect to completely neutralize each other and make 0.0341 mol H₂O.

Combining our two calculations will give us how much heat is released per mol of H₂O:

1,942.21 J/0.0341 mol H₂O) = 56,956.4 J/mol

= 57.0 KJ/mol H₂O (rounding to correct Sig Figs)