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

Post by Dhamnekar Winod »

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?
Last edited by Dhamnekar Winod on Sat Jul 03, 2021 9:44 am, edited 1 time in total.
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]
Orcio_Dojek
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Re: Enthalpy of reaction using one equation and its enthalpy

Post by Orcio_Dojek »

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?
No, it is obvious wrong.
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Dhamnekar Winod
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Re: Enthalpy of reaction using one equation and its enthalpy

Post by Dhamnekar Winod »

For readers and viewers , detailed answer to this question is not transferrable
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]
Warner Mortensen
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Re: Enthalpy of reaction using one equation and its enthalpy

Post by Warner Mortensen »

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
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Dhamnekar Winod
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Re: Enthalpy of reaction using one equation and its enthalpy

Post by Dhamnekar Winod »

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]
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