heats of reaction write-up

Heats of Reaction









Michael DiPietro

Thurs PM

Partners: George Duryea & Elisabeth Cianciola

Acid and Base Assignment: HONH3Cl and KOH

2mL, 4mL, 6mL, 8mL

RESULTS



Concentration of HONH3Cl: 0.886M

Concentration of KOH: 0.860 M



Raw Data for neutralization reaction:

Volume (acid) (mL) Volume (base) (mL) Ti (acid) (ºC) Ti (base) (ºC) Tf (ºC)



2 10 23.72 22.90 23.40

4 10 22.06 23.05 23.84

6 10 22.07 22.19 24.40

8 10 22.12 22.92 24.34



Calculated Data for neutralization reaction:

Moles of acid neutralized Ti (weighted T (ºC) Total Mass of Solution q(rxn)

(mol) average) (ºC) (g) (ºC)



0.002 23.03 0.37 12 -19

0.004 22.76 1.1 14 -64

0.005 22.14 2.3 16 -135

0.007 22.56 1.8 18 -154



Sample calculations:

Moles acid neutralized: 2mL * (0.886 mol/1000mL) = .001772 mol  .002 mol

HONH3Cl



Weighted Average: [(23.72(2.0 mL) + (22.90)(10.0 mL)]/12 mL = 23.03



T: 23.40-23.03= 0.37



Total Mass of Solution: 0.998 * 12mL = 11.97g  12 g



q(rxn): 12g * 4.18 * 0.37 = 18.55  19 qsol  -19 q(rxn)

Graph 1:



Moles of Acid Neutralized v. q(rxn)



40

20

0

-20 0 0.002 0.004 0.006 0.008



-40

q(rxn)









-60 q(rxn)

-80 Linear (q(rxn))

-100

y = -24240x + 12.863

-120

R2 = 0.9164

-140

-160

-180

Moles of Acid Neutralized





The ∆H seems to change at an inconsistent rate per mole of acid neutralized.



Pooled Class Data:

Reaction

HNO3 + NaOH -51.45

HNO3 + NaOH -62.41

HCL + NaOH -58.25

HCl + KOH -54.13

HONH3Cl + NaOH -23.02



DISCUSSION:

From what I can determine from my data, the enthalpy change is an extensive property

due to the fact that all of the ∆H are negative, instead of positive. This shows that the

reaction is releasing energy, instead of gaining it.



Using the class data it can also be concluded that the enthalpy change is related to the

identity of the bases and acids as different combinations result in more or less heat being

lost.