## 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

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℃

Last edited by Dhamnekar Winod on Sat Jul 03, 2021 9:44 am, edited 1 time in total.
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Orcio_Dojek
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Joined: Sun Jun 20, 2021 7:55 am

### Re: Enthalpy of reaction using one equation and its enthalpy

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℃ 