Q-Hess’s Law

Prem SattsangiCopyright 2007

#2-Hess’s Law Study

• Please have your pencil and paper and BLB text book as you attempt these problem.

• Write your response.

• Press enter to check the correct answer.

#3-Definitions BLB-Page 170

Please Define:

SYSTEM:Portion of the Universe under study.

SURROUNDINGS: Rest of the Universe.

#4-Definitions BLB-Page 174

Please Define:

ENDOTHERMIC:Process in which the system Absorbs Heat.

EXOTHERMIC: Process in which the system Evolves Heat.

p. 175 Fig. 5.8p. 178 Fig. 5.12

#5-Definitions BLB-Page 183

Please Define:

HEAT CAPACITY:Amount of Heat required to raise the temperature of an

object by 1 oC.

SPECIFIC HEAT : HEAT CAPACITY per gram of a substance. ORAmount of Heat required to raise the temperature of 1g of

a substance by 1 oC.

#6-Definitions

Please Define: HESS’S LAW: (p. 189) If a reaction is carried out in a series of steps, H for the overall

reaction will equal the sum of enthalpy changes for individual steps. PRACTICE: SE 5.8

STANDARD STATE : (p. 191) Physical State (s, l, or g) of a substance under STANDARD

conditions, 1 atm and 25oC STANDARD ENTHALPY OF FORMATION:

Hform of 1 mol of a compound starting from its elements with all substances under STANDARD conditions.

#7 Hess’s Law calculations (p. 189)

• Calculate Hform for C3H8 [A: -106 kJ]• REACTANTS PRODUCTS• 3C(s) + 4H2(g) C3H8 (g) FINAL EQUATION

• GIVEN:(a) C(s) + O2(g) CO2 (g) -394 kJ(b) H2(g) + ½ O2(g) H2O (l) -286 kJ(c) C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O (l) -2220.0 kJ

Rewrite each equation to:(i) MATCH the REACTANTS and PRODUCTS in the FINAL EQUATION.(ii) If you reverse the equation, change the sign of H.

(a) 3C(s) + 3O2(g) 3 CO2 (g) -394 kJ x 3 (to get 3C)(b) 4H2(g) + 2 O2(g) 4 H2O (l) -286 kJ x 4 (to get 4 H2)(c) 3 CO2(g) + 4 H2O (l) C3H8(g) + 5 O2(g) +2220.0 kJ

Add the equations to obtain the FINAL EQUATION. Calculate the kJ.

#8- Calculating”Q/g”, p. 183, SE 5.5

• Q = m(H2O) x s (H2O) x T Eq. 5.22

• [s (H2O) = 4.184 J/goC, m (water only) ]

• Calculate Hsoln./g for the U.K.#1

• Wt. Of sample = 2.05 g• Amt of water = 100 mL (D = 1g/mL)T = 1.65oC

• Q = 100 g x 4.184 J/goC x 1.65oC = 690 J

• J/g sample = 690 J/2.05 g = 336.6 J/g

#9- Calculating” Hsoln. /mol”

Hsoln./g for the U.K.#1 was = 336.6 J/g

• Calculate Hsoln. /mol in kJ, if the U.K.#1 was NH4NO2. [Atomic weights: N(14.0), H(1.01), O(16.0)]

• FW: NH4NO2 = 28.0 + 4.04 + 32.0 = 64.0 g

Hsoln. = 336.6 J x 64.0 g x 1kJ___ = 21.54 kJ/mol mol g mol 1000J

#10-Calculating Q/mol, Heat of Neutralization of solutions

• On mixing 60.0 mL each of 0.50 M HCl(aq) and 0.50 M KOH(aq) the temperature rose by 7.25 oC. Calculate H per mol KOH.

Total volume of Aq. Solutions = 60.0 mL + 60.0 mL= 120.0 mL

•Hence: Q = 120 g x 7.25 oC x 4.184 J = 3,640.1 J• g oC

•Mol KOH present in 60.0 mL of 0.50 M solution(aq): •n = M x V or 0.50 mol x 60.0 mL x 1 L = 0.030 mol • L 1x103 mL

•Q = 3,640.1 J x ( 1 kJ) = 121 kJ•Mol 0.030 mol (1000J) mol

#11 Neutralization reactions Role of a LIMITING REAGENT

• If you used 50.0 mL each of the following solutions, Heat of Neutralization of sets (b) and (c) will be

• (I) Same as (a), (ii) more, (iii) Less LIMITING REAGENT

• (a) 0.5 M HCl + 0.5 M NaOH NONE• (b) 0.4 M HCl + 1.0 M NaOH Reagent with lower “M”• (c) 1.3 M HCl + 0.6 M NaOH Reagent with lower “M”

• Consider the role of LIMITING REAGENT• (c) 0.6 M NaOH is less than 1.3 M HCl. • It is the LIMITING REAGENT

• Comparing it to “0.5 M NaOH in (a),” this is more. • Heat of Neutralization of set (C) will be more.