Find the enthalpy of neutralization of 75.8 mL 1.00M aqueous sodium hydroxide with 76.0 mL 1.00M hydrochloric acid if after mixing the two solutions the temperature went up 5.7°C. Report the amount of the limiting reagent: mmol Report the enthalpy of reaction if the heat capacity of the styrofoam cup is 62.3 J/K kJ/mol
Find the enthalpy of neutralization of 75.8 mL 1.00M aqueous sodium hydroxide with 76.0 mL 1.00M hydrochloric acid if after mixing the two solutions the temperature went up 5.7°C. Report the amount of the limiting reagent: mmol Report the enthalpy of reaction if the heat capacity of the styrofoam cup is 62.3 J/K kJ/mol
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter8: Thermochemistry
Section: Chapter Questions
Problem 11QAP: When one mol of KOH is neutralized by sulfuric acid, q=56 kJ. (This is called the heat of...
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![Pre-lab: Theory
The heat of neutralization of solid metal hydroxide with an aqueous solution of an acid
might be determined directly in a one-step reaction
МОН (:) + H M'+ Hә0(1)
or indirectly using a two-step reaction
MOH(s)
- Mt + OH
OH +H* → H2O(1)
Based on the net ionic equation of the reaction, the theoretical enthalpy of reaction,
AHneo kJ/mol], can be found from the enthalpies of formation as
AHheo =
AH; [H2O(1)] + AH; [M* (aq)] – AH¡ [MOH(s)]
Based on the measured change in temperature, AT[K], and the heat capacity of the cup,
Ccup[J /K], the experimental enthalpy of reaction, AHrp[kJ/mol], can be found as
= - (Su Vtot + Ccup) AT/nL.R.,
where Vtot (mL] is the total volume of the solution, nL.R.[mmol] is the amount of the
limiting reagent, and specific heat of water is s, = 4.18 JmL-'K-!
The percent error can be found according to the formula
ΔΗ
'exp
AHiheo
-AHiheo
x 100%
%E =
Using the values for standard enthalpies of formation from the table 8-1 find the enthalpies
for the reactions in the following questions.
Question 1
Question 2
Find the enthalpy of neutralization of 75.8 mL 1.00M aqueous sodium hydroxide
with 76.0 mL 1.00M hydrochloric acid if after mixing the two solutions the
temperature went up 5.7°C. Report the amount of the limiting reagent:
mmol
Report the enthalpy of reaction if the heat capacity of the styrofoam cup is 62.3 J/K
kJ/mol](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1a88c37e-6615-4150-bb38-f5623b75f939%2Fb90304d2-a49a-4205-8037-d15175704356%2Fbiegd7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Pre-lab: Theory
The heat of neutralization of solid metal hydroxide with an aqueous solution of an acid
might be determined directly in a one-step reaction
МОН (:) + H M'+ Hә0(1)
or indirectly using a two-step reaction
MOH(s)
- Mt + OH
OH +H* → H2O(1)
Based on the net ionic equation of the reaction, the theoretical enthalpy of reaction,
AHneo kJ/mol], can be found from the enthalpies of formation as
AHheo =
AH; [H2O(1)] + AH; [M* (aq)] – AH¡ [MOH(s)]
Based on the measured change in temperature, AT[K], and the heat capacity of the cup,
Ccup[J /K], the experimental enthalpy of reaction, AHrp[kJ/mol], can be found as
= - (Su Vtot + Ccup) AT/nL.R.,
where Vtot (mL] is the total volume of the solution, nL.R.[mmol] is the amount of the
limiting reagent, and specific heat of water is s, = 4.18 JmL-'K-!
The percent error can be found according to the formula
ΔΗ
'exp
AHiheo
-AHiheo
x 100%
%E =
Using the values for standard enthalpies of formation from the table 8-1 find the enthalpies
for the reactions in the following questions.
Question 1
Question 2
Find the enthalpy of neutralization of 75.8 mL 1.00M aqueous sodium hydroxide
with 76.0 mL 1.00M hydrochloric acid if after mixing the two solutions the
temperature went up 5.7°C. Report the amount of the limiting reagent:
mmol
Report the enthalpy of reaction if the heat capacity of the styrofoam cup is 62.3 J/K
kJ/mol
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