Question

In a constant-pressure calorimeter, 70.0 ml of 0.310 M Ba(OH)_2 was added to 70.0 ml of 0.620 M HCl. The reaction caused the temperature of the solution to rise from 21.27°C to 25.49°C. If the solution has the same density and specific heat as water, what is ΔH for this reaction?

a) -4.86 kJ/mol
b) -2.43 kJ/mol
c) 4.86 kJ/mol
d) 2.43 kJ/mol

Answer 1

ΔH for the Ba(OH)2 and HCl reaction can be calculated using the specific heat formula for water, identifying the limiting reactant, and then relating the heat released to the moles of the limiting reactant.

Step-by-step explanation:

To determine the ΔH for the reaction between Ba(OH)2 and HCl, we can use the formula q = mcΔT, where q is the heat absorbed or released, m is the mass of the solution, c is the specific heat capacity, and ΔT is the change in temperature. Since the reaction occurred in a constant-pressure calorimeter, ΔH is equal to the heat q divided by the number of moles of the limiting reactant. The reaction is:

Ba(OH)2 (aq) + 2HCl (aq) → BaCl2 (aq) + 2H2O (l)

After determining the moles of each reactant, we identify HCl as the limiting reactant because it has fewer moles available for the reaction. By calculating the total mass of the solution (140.0 mL × 1.00 g/mL = 140.0 g) and using the specific heat capacity of water (4.184 J/g°C), we can find q and then calculate ΔH per mole of HCl.