PreLab questions for Hess\'Law Lab CHEM 1211L Lab Manual-Page 113 Hess\' Law Pre

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PreLab questions for Hess'Law Lab

CHEM 1211L Lab Manual-Page 113 Hess' Law Pre-Lab Exercises Name: Mesut R Before coming to lab, read the sections on calorimetry and Hess's Law in your textbook 1. Define the following terms ci TMd_rryn ,dymo~?4, OS-t-efur.ch/anig a, state function C4.de, ?arre" hde, 5hdeqvu..J. Jtes ?ka+ A, , riso-? trr c. enthalpy The chamyen 2. Write the balanced chemical equation for A the reaction of magnesium with aqueous hydrochloric acid B. the reaction of magnesium oxide with aqueous hydrochloric aciod. C. the formation of liquid water from gaseous hydrogen and gaseous oxygen. D. the combustion of magnesium metal to form 1 mole of magnesium oxide

Explanation / Answer

3a) The three equations are given as

Mg (s) + 2 HCl (aq) --------> H2 (g) + MgCl2 (aq); ?H1 ……….(A)

MgO (s) + 2 HCl (aq) ---------> H2O (l) + MgCl2 (aq); ?H2 ……..(B)

H2 (g) + ½ O2 (g) --------> H2O (l); ?H3 ……..(C)

Add (A) and (C) and subtract (B) to get

Mg (s) + 2 HCl (aq) + H2 (g) + ½ O2 (g) – MgO (s) – 2 HCl (aq) ---------> H2 (g) + MgCl2 (aq) + H2O (l) – H2O (l) – MgCl2 (aq)

Cancel out common terms from both sides and get

Mg (s) + ½ O2 (g) – MgO (s) ----------> 0

Add MgO to both sides and get

Mg (s) + ½ O2 (g) ---------> MgO (s); ?H4

The above represents equation (D) in (2).

3b) The enthalpy (?H) terms are additive; hence, we have

?H4 = ?H1 + ?H3 – ?H2

4a) The heat absorbed by water is given as

Heat absorbed = (mass of water)*(specific heat capacity of water)*(change in temperature)

= (mass of water)*(4.184 J/g.°C)*(final temperature – initial temperature)

= (450 g)*(4.184 J/g.°C)*(95 – 15)°C = 150624 J = (150624 J)*(1 kJ/1000 J) = 150.624 kJ (ans).

b) The system is the water. Since the water absorbs heat, by convention, the change in enthalpy of the system is assigned positive.

The enthalpy change of the system is the heat absorbed divided by the moles of water.

We have 450 g water. The molar mass of water is (2*1 + 1*16) g/mol = 18 g/mol.

Therefore, 450 g water = (450 g)*(1 mole/18 g) = 25 mole.

Enthalpy change of the system = heat absorbed/moles of water = (150.624 kJ)/(25 mole) = 6.02496 kJ/mol ? 6.025 kJ/mol (ans).

c) Assume that the system and the surroundings form an isolated system; therefore, there is no heat or energy exchange with the surroundings. Hence, the enthalpy change of the isolated system is zero, i.e,

(Enthalpy change of the system) + (Enthalpy change of the surroundings) = 0

====> (+6.025 kJ/mol) + (Enthalpy change of the surroundings) = 0

====> Enthalpy change of the surroundings = -6.025 kJ/mol (ans).

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