An Ice Calorimeter Determination of Reaction Enthalpy

1564 Words7 Pages
Thermochemistry: An Ice Calorimeter Determination of Reaction
Enthalpy
D. F. Nachman
6/23/2010
Abstract: An ice calorimeter was used to study the reaction of magnesium metal and 1.00M sulfuric acid solution: Mg(s) + H2SO4(aq) →MgSO4(aq) + H2(g). We found the experimental molar enthalpy of reaction to be ΔH = –355 ± 17 kJ/mol at 0°C, 24% lower than the textbook value of ΔH° = –466.9 kJ/mol, reported at 25°C.
Introduction
Whether a chemical reaction occurs spontaneously or is driven by an outside force, it almost always exchanges energy with the surroundings. Energy exchange can occur as work or as heat flow. When a reaction occurs under constant-pressure conditions, we call the heat portion of the energy exchanged the enthalpy change of
…show more content…
Calculations
The reaction between magnesium metal and sulfuric acid released heat, which melted some of the ice in the calorimeter. Because ice takes up more volume per gram than liquid water, this melting produced a drop in pipette readings. For each gram of ice that melted, the volume change was calculated from the density values of water and ice at 0 °C: dice 0.9157 g
1 mL
, so the specific volume of ice =
1 mL
0.9157 g
1.0920 mL ice
1 g H2O dliquid 0.9998 g
1 mL
, so the specific volume of liquid =
1 mL
0.9998 g
1.0002 mL
1 g
The difference between the two specific volume values gives the volume change per gram of melting ice: 0.0918 mL/g.
The estimated total volume change is shown in Figure 1. Because the slope of the data trend after the reaction did not match the slope of the data trend before the reaction, using the equations of the trend lines gave a range of ΔV values:
ΔVmin = 0.8989–0.5914+[630*(-0.00005+0.0003)]=0.3075+[630*0.00025]=0.465 mL
ΔVavg = 0.8989–0.5914+[720*(-0.00005+0.0003)]=0.3075+[720*0.00025]=0.488 mL
ΔVmax = 0.8989–0.5914+[810*(-0.00005+0.0003)]=0.3075+[810*0.00025]=0.510 mL
These volume changes correspond to the following masses of ice melt:
0.465 mL melt 1 g ice
0.0918 mL melt
= 5.07 g ice minimum
0.488 mL melt 1 g ice
0918 mL melt
5.32 g ice average
0.510 mL melt 1 g ice
0.918 mL melt
5.56 g ice maximum
Knowing

More about An Ice Calorimeter Determination of Reaction Enthalpy

Open Document