Enthalpy Lab Report Essay

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Abstract
This lab is performed in order to determine the total energy in a reaction between zinc and hydrochloric acid. The reaction is done twice, once to measure the heat of the reaction and again to determine the work done in the system. This is because Enthalpy equals heat plus work (∆H= ∆E+W). Heat and work can be broken down further into separate components so the equation used in lab is ∆H=mc∆T + PV. Many calculations are used in the lab to find out what cannot be measured directly (ex: volume). After all the calculations were complete it was shown to have a very small percent error.

Introduction: The theory behind this experiment is the heat of a reaction (∆E) plus the work (W) done by a reaction is equal to …show more content…
This will determine the enthalpy for the reaction. (Procedure is derived from the combined thinking of Austin Slutsky and Max McNutt).
Data and Results:
Substance Mass (g) Ti (⁰C) Tf (⁰C) ΔT(⁰C) P(mmHg) V cm3
HCl 48.317g 22.9 40.4 17.8 28.388
Zn 3.159g 22.9 40.4 17.8 28.388
H2 141cm3
Find ml of HCl needed to react with 3.159g of Zn
3.159gZn x (1 mole Zn/65.38gZn) x (2 moles HCl/1 mole Zn) x (1000ml HCl/2 moles Zn) ml HCl=48.317ml Find Volume of H2
C=2πr
20.29cm=2πr
R=3.229cm
Area=(4/3) πr3
A=(4/3) π3.229
A=141cm3

Find ΔT
ΔT=(Tfinal-Tinitial)
ΔT=(40.4C-22.9C)
ΔT=17.8C

Find ΔH
ΔE=mcΔT
W=PV
ΔH=mcΔT+PV
ΔE=(48.317g)(4.18J/gC)917.5C)
ΔE=73633J
W=(28.388mmHg)(141cm3)
W=83.389J
ΔH=[(48.317g)(4.18J/gC)917.5C)/.048molZn]+[ (28.388mmHg)(141cm3)/.048molZn]
ΔH=73633J+83.389J
ΔH=-157KJ/mol
Percent Error
(-157J)+(159.88J)/-159.88 x 100 =1.8%error
Discussion and Conclusions: This experiment was initiated when zinc was put into a flask containing hydrochloric acid. When the two come into contact hydrogen gas was released as zinc chloride was formed. Violent bubbling happened as the reaction took place, these bubbles

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