How much heat is produced when 100 mL of 0.250 M HCl (density, 1.00 g/mL) and 200 mL of 0.150 M NaOH (density, 1.00 g/mL) are mixed?
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) ∆H°₂₅°c = −58 kJ If both solutions are at the same temperature and the heat capacity of the products is 4.19 J/g℃, how much will the temperature increase? What assumption did you make in your calculation?
My answer: Reaction given produces 58 kJ of heat with 1 mol of NaOH. We have 0.150 M of NaOH (limiting reagent) so it will produce 0.150 M × -58 kJ/1.0 M= -8.7 kJ of heat.
We know that q=cm∆T, substituting the values we have , in the equation -8.7 kJ = 4.19 J/g℃ × 300 g × ∆T, ⇒ ∆T= 6.9℃
Is this answer correct?
Enthalpy of reaction using one equation and its enthalpy
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- Dhamnekar Winod
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Enthalpy of reaction using one equation and its enthalpy
Last edited by Dhamnekar Winod on Sat Jul 03, 2021 9:44 am, edited 1 time in total.
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Re: Enthalpy of reaction using one equation and its enthalpy
No, it is obvious wrong.How much heat is produced when 100 mL of 0.250 M HCl (density, 1.00 g/mL) and 200 mL of 0.150 M NaOH (density, 1.00 g/mL) are mixed?
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) ∆H°₂₅°c = −58 kJ If both solutions are at the same temperature and the heat capacity of the products is 4.19 J/g℃, how much will the temperature increase? What assumption did you make in your calculation?
My answer: Reaction given produces 58 kJ of heat with 1 mol of NaOH. We have 0.150 M of NaOH (limiting reagent) so it will produce 0.150 M × -58 kJ/1.0 M= -8.7 kJ of heat.
We know that q=cm∆T, substituting the values we have , in the equation -8.7 kJ = 4.19 J/g℃ × 300 g × ∆T, ⇒ ∆T= 6.92℃
Is this answer correct?
- Dhamnekar Winod
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Re: Enthalpy of reaction using one equation and its enthalpy
For readers and viewers , detailed answer to this question is not transferrable
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Re: Enthalpy of reaction using one equation and its enthalpy
Finding Moles via
0.250 M ∗ 100 mL / 1000 mLL = 0.0250 mol HCl
0.150 M∗200 mL / 1000 mLL = 0.0300 mol NaOH
Assuming that there is limiting reactant
ΔH∘298 = q = 0.0250 ∗ − 58 kJ ∗ 1000 J / KJ = − 1450 J
Assuming that the heat is fully transferred
− qrxn = qcal = 1450J
Using equation
q = mΔT
Finding Mass
(100 mL ∗ 1 gm / L) + (200 m L ∗ 1 gm / L) = 300 g
1450 J = (300 g) (4.19 J / g C∘) ΔT
Δ T = 1450 J / (300 g) (4.19 JgC∘) ≈ 1.1535 C∘
Hence, temperature of the solution will increase by 1.1535 degrees
0.250 M ∗ 100 mL / 1000 mLL = 0.0250 mol HCl
0.150 M∗200 mL / 1000 mLL = 0.0300 mol NaOH
Assuming that there is limiting reactant
ΔH∘298 = q = 0.0250 ∗ − 58 kJ ∗ 1000 J / KJ = − 1450 J
Assuming that the heat is fully transferred
− qrxn = qcal = 1450J
Using equation
q = mΔT
Finding Mass
(100 mL ∗ 1 gm / L) + (200 m L ∗ 1 gm / L) = 300 g
1450 J = (300 g) (4.19 J / g C∘) ΔT
Δ T = 1450 J / (300 g) (4.19 JgC∘) ≈ 1.1535 C∘
Hence, temperature of the solution will increase by 1.1535 degrees
- Dhamnekar Winod
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Re: Enthalpy of reaction using one equation and its enthalpy
Your answer looks correct. Thank you.
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]