Calculate the expected ΔHrxn using the ΔHf values for the reactants andproducts in the Reaction. (Hint: this is actually much easier if you writethe net reaction first and then calculate ΔHrxn for the net.)HCl(aq) + NaOH(aq) -> H2O(l) + NaCl(aq)i attached the table for the problem and my work up to this point. plz only solve for trial one. i am wondering how delta H will change when the concentration of HCl changes. Note the lab question isHow does the concentration of acid affect the heat of reaction (ΔHrxn)between a strong acid (HCl) and a strong base(NaOH)?this is for reference Table 1: Trial DataMass of cups (g)Volume of HCI (mL)Volume of NaOH (mL)3.5095050NaOH concentration (M)0.955Initial temperature of compounds (°C)17.75Trial #1Trial #2Trial #3HCl concentration (M)0.9230.6920.462Final temperature of compounds (°C)Mass of solution (kg)24.322.821.20.1032550.101990.102134AT (°C)6.555.053.35Energy/q (kJ)Change in enthalphy/AHrxn (kJ/mol)3.572.151.43-55.83-55.83-55.83 a) (msolution) = total mass(msolution) = 106.764g – 3.509g = 103.255gb) AT = Tt - TiAT = 24.30°C – 17.75°C = 6.550°Cc) q = m*Cs* ATq = 0.103255kg * 4.184kJ/K*6.550°C = 3.56966kJd) n(HCI) = (0.923mol/L)(50mL)(1L/1000mL) = 0.046moln(NaOH) = (0.955mol/L)(50mL)(1L/1000OmL) = 0.048molAccording to the balanced equation: HCI(aq)+NaOH(aq)→NaCI(aq)+H20(1), the stoichiometric ratio between the moles of HCl and the productmoles of H20 produced is of a 1:1 ratio, thus 0.046mol of HCI will reactwith 0.048mol of NaOH to produce 0.046mol of H2Omass of cups

In an neutralisation reaction between a strong acid HCl and strong base NaOH, the acid give H+ ions…

Answered: Calculate the expected ΔHrxn using the…

World of Chemistry, 3rd edition

3rd Edition

ISBN:9781133109655

Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste

Chapter10: Energy

Section: Chapter Questions

Problem 8A

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Calculate the expected ΔHrxn using the ΔHf values for the reactants and
products in the Reaction. (Hint: this is actually much easier if you write
the net reaction first and then calculate ΔHrxn for the net.)

HCl(aq) + NaOH(aq) -> H2O(l) + NaCl(aq)

i attached the table for the problem and my work up to this point. plz only solve for trial one. i am wondering how delta H will change when the concentration of HCl changes. Note the lab question is

How does the concentration of acid affect the heat of reaction (ΔH_rxn)between a strong acid (HCl) and a strong base(NaOH)?

this is for reference

Table 1: Trial Data
Mass of cups (g)
Volume of HCI (mL)
Volume of NaOH (mL)
3.509
50
50
NaOH concentration (M)
0.955
Initial temperature of compounds (°C)
17.75
Trial #1
Trial #2
Trial #3
HCl concentration (M)
0.923
0.692
0.462
Final temperature of compounds (°C)
Mass of solution (kg)
24.3
22.8
21.2
0.103255
0.10199
0.102134
AT (°C)
6.55
5.05
3.35
Energy/q (kJ)
Change in enthalphy/AHrxn (kJ/mol)
3.57
2.15
1.43
-55.83
-55.83
-55.83

a) (msolution) = total mass
(msolution) = 106.764g – 3.509g = 103.255g
b) AT = Tt - Ti
AT = 24.30°C – 17.75°C = 6.550°C
c) q = m*Cs* AT
q = 0.103255kg * 4.184kJ/K*6.550°C = 3.56966kJ
d) n(HCI) = (0.923mol/L)(50mL)(1L/1000mL) = 0.046mol
n(NaOH) = (0.955mol/L)(50mL)(1L/1000OmL) = 0.048mol
According to the balanced equation: HCI(aq)+NaOH(aq)→NaCI(aq)+H
20(1), the stoichiometric ratio between the moles of HCl and the product
moles of H20 produced is of a 1:1 ratio, thus 0.046mol of HCI will react
with 0.048mol of NaOH to produce 0.046mol of H2O
mass of cups

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