In a constant‑pressure calorimeter, 55.0 mL of 0.340 M Ba ( OH ) 2 was added to 55.0 mL of 0.680 M HCl . The reaction caused the temperature of the solution to rise from 21.47 ∘ C to 26.1 ∘ 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 H 2 O produced)? Assume that the total volume is the sum of the individual volumes.

ΔH  = 57.0 kJ/mol

Explanation:

Step 1: Data given

Volume of a 0.340 M Ba(OH)2 solution = 55.0 mL = 0.0550 L

Volume of a 0.680 M HCl = 55.0 mL = 0.0550 L

The reaction caused the temperature of the solution to rise from 21.47  °C to 26.1°C

Density of water = 1.00 g/mL

Specific heat of water = 4.184 J/g°C

Step 2: The balanced equation

Ba(OH)2 + 2 HCl → BaCl2 + 2 H2O

Step 3: Calculate moles

Moles = molarity * volume

Moles Ba(OH)2 = 0.340 M * 0.055 L

Moles Ba(OH)2 = 0.0187 moles

Moles HCl = 0.680 M * 0.055 L

Moles HCl = 0.0374 moles

Step 4: Calculate limiting reactant

For 1 mol Ba(OH)2 we need 2 moles HCl to produce 1 mol BaCl2 and 2 moles H2O

For 0.0187 moles Ba(OH)2 we need 0.0374 moles HCl

Both reactants will be totally consumed.

Neither is in excess.

Step 5: Calculate moles H2O

For 1 mol Ba(OH)2 we need 2 moles HCl to produce 1 mol BaCl2 and 2 moles H2O

There will be produced 0.0374 moles H2O

Step 6: Calculate heat absorbed by the solution

Q = m*c* ΔT

⇒Q = the heat absorbed by the solution = TO BE DETERMINED

⇒m =the mass = 110 mL * 1.00 g/mL = 110 grams

⇒c= the specific heat of water = 4.184 J/g°C

⇒ΔT = the change of temperature = T2 - T1 = 26.1 °C - 21.47 °C = 4.63 °C

Q = 110 * 4.184 * 4.63

Q = 2130.9 J is absorbed by the solution

Step 7: Calculate  ΔH for this reaction

ΔH = 2130.9 J / 0.0374 moles H2O

ΔH = 56976 J/mol = 57.0 kJ/mol H2O

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