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 21.80 °C to 26.02 °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)? Assume that the total volume is the sum of the individual volumes.

Answer 1

Ans. # Step 1: Balanced reaction: Ba(OH)₂ + 2 HCl(aq) -----> BaCl₂(aq) + 2 H₂O(l)

# Moles of Ba(OH)₂ taken = Molarity x Vol. in liters = 0.310 M x 0.055 L = 0.01705 mol

            Moles of OH^- = 2 x Moles of Ba(OH₂) = 2 x 0.01705 mol = 0.0341 mol

# Moles of HCl taken = 0.620 M x 0.055 L = 0.0341 mol

Since moles of OH^- is equal to that of H⁺ or HCl, the acid and base are completely neutralized by each other.

            OH- + H₃O⁺ ---> 2 H₂O

Moles of H₂O formed = 2 x Moles of OH^- = 2 x 0.0341 mol = 0.0682 mol

# Step 2:

Note: The dHmolar is around ½ of the actual value. Please cross-check if there has been an error with the experimental values.