A student, using an ice calorimeter, studies the enthalpy of reaction of the fol

Question

A student, using an ice calorimeter, studies the enthalpy of reaction of the following exothermic chemical change: Mg(s) + 2H+(aq) ? ? Mg2+(aq) + H2(g) From data collected, it was determined that a magnesium sample having a mass of 0.0689 grams would react with excess HCl and melt the ice in the calorimenter that resulted in a volume change of 0.310 ml. Determine the following by referring to your laboratory procedures. Give special consideration to the section dealing with calculation. Calculate a value for the enthalpy change H during the reaction between HCl and Mg?Mg(s) + 2H+ (aq)? Mg+2 (aq) + H2 (g) ?1H (1)
MgO(s) + 2H+ (aq) ? Mg+2 (aq) + H2O (?) ?2H (2) A value for the enthalpy of formation of water is available from the literature.
H2 (g) +1?2O2 (g) ? H2O(?) ?3H (3) These three reactions can be combined to yield the formation reaction of magnesium oxide.
Mg(s) + 1?2O2 (g) ? MgO(s)
?4H (4)

Explanation / Answer

Well first we need to find the number of grams of ice melted. We don't know if the 0.300 ml volume change is an increase or a decrease so we will have to just pick one and correct if negative.

V_else = Volume of everything except ice or water
V_ice = Volume occupied by ice
V_water = Volume occupied by water

V_else + V_ice = V_else + V_water + 0.300 mL

V_else drops out of both sides
V = density*m

density_ice*m = density_water*m + 0.300 mL
(density_ice - density_water)*m = 0.300 mL

density of ice = 0.917 g/mL
density of water = 1.0 g/mL

m = 0.300 mL /(0.917 g/mL -1.0 g/mL)
m = 3.61 g

Now to find the heat of formation. We will assume that no heat escaped and none was added and that the heat of formation is equal to the heat that melted the 3.61 g of water.

q_form = q_melt.
DHf(Mg2+)*m_Mg2 = DHm(H2O)*m_H2O
DHf(Mg2+) = DHm(H2O)*m_H2O/m_Mg2

The heat of melting for water, DHm(H2O), is 334 J/g.
m_H2O = 3.61 g
m_Mg2 = 0.0711 g

DHf(Mg2+) = 334 J/g *3.61 g / 0.711 g
DHf(Mg2+) = 16,958 J/g

Use the molecular weight of Mg to convert to mol, 24.305 g/mol, and use the fact that 1 calorie = 4.1868 joules.

DHf(Mg2+) = 16,958 J/g * (1cal / 4.1868 J) * (24.305 g/ mol)
DHf(Mg2+) = 98,444 cal / mol

DHf(Mg2+) = 98.4 kcal/mol

On to the sign. I always left the sign for last. It was easier for me that way. The reaction gives off heat and hence loses it which makes it negative.

DHf(Mg2+) = -98.4 kcal/mol

I just looked at my Physical Chemistry book and it says -111.58 kcal/mol so I think it's close enough.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.