1. The Hrxn for the following reaction is −56.27 kJ. Calculate the amount of heat required to convert 1.00 g of calcium hydroxide to calcium oxide.
CaO (s) + H2O (l)Ca(OH)2 (s)
2. A 0.500 L sample of 0.200 M NaCl (aq) is added to 0.500 L of 0.200 M AgNO3 (aq) in a calorimeter with a heat capacity of 4.60 × 103 J.K−1. The observed temperature change is +1.423 K degrees.
Calculate the value of the heat of reaction for the following:
AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq)
Chemistry Help!!?
2020-05-17 6:35 pm
回答 (1)
2020-05-17 7:03 pm
1.
Molar mass of Ca(OH)₂ = (40.1 + 16.0×2 + 1.0×2) g/mol = 74.1 g/mol
CaO(s) + H₂O(ℓ) → Ca(OH)₂(s) ΔHrxn = -56.27 kJ
Ca(OH)₂(s) → CaO(s) + H₂O(ℓ) ΔHrxn = -(-56.27 kJ) = +56.27 kJ
When 74.1 g of Ca(OH)₂ reacts, 56.27 kJ of heat is required.
When 1.00 g of Ca(OH)₂ reacts, heat required = (56.27 kJ) × (1/74.1) = 0.759 kJ
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2.
Density of the solution ≈ Density of water = 1.00 g/mL
Mass of the final solution = [(0.500 + 0.500) L] × (1000 mL/L) × (1.00 g/mL) = 1000 g
Specific heat capacity of the solution, c ≈ Specific heat capacity of water = 4.184 J g⁻¹ C⁻¹)
Heat absorbed by the solution = m c ΔT = 1000 × 4.184 × 1.423 J = 5954 J
Heat absorbed by the calorimeter = (4.60 × 10³) × 1.423 = 6546 J
Heat released by the reaction = (5954 + 6546) J = 12500 J = 12.5 kJ
AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq) ΔHrxn < 0 as temperature increases
Moles of AgNO₃ reacted = Moles of NaCl reacted = 0.200 × 0.500 mol = 0.1 mol
ΔHrxn = (-12.5/0.1) kJ = -125 kJ
Molar mass of Ca(OH)₂ = (40.1 + 16.0×2 + 1.0×2) g/mol = 74.1 g/mol
CaO(s) + H₂O(ℓ) → Ca(OH)₂(s) ΔHrxn = -56.27 kJ
Ca(OH)₂(s) → CaO(s) + H₂O(ℓ) ΔHrxn = -(-56.27 kJ) = +56.27 kJ
When 74.1 g of Ca(OH)₂ reacts, 56.27 kJ of heat is required.
When 1.00 g of Ca(OH)₂ reacts, heat required = (56.27 kJ) × (1/74.1) = 0.759 kJ
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2.
Density of the solution ≈ Density of water = 1.00 g/mL
Mass of the final solution = [(0.500 + 0.500) L] × (1000 mL/L) × (1.00 g/mL) = 1000 g
Specific heat capacity of the solution, c ≈ Specific heat capacity of water = 4.184 J g⁻¹ C⁻¹)
Heat absorbed by the solution = m c ΔT = 1000 × 4.184 × 1.423 J = 5954 J
Heat absorbed by the calorimeter = (4.60 × 10³) × 1.423 = 6546 J
Heat released by the reaction = (5954 + 6546) J = 12500 J = 12.5 kJ
AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq) ΔHrxn < 0 as temperature increases
Moles of AgNO₃ reacted = Moles of NaCl reacted = 0.200 × 0.500 mol = 0.1 mol
ΔHrxn = (-12.5/0.1) kJ = -125 kJ
收錄日期: 2021-04-12 12:52:47
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