Can you help you number 1, 2 nad 3 SJCC Instructor: Lazik Chemistry 1A Practice

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Can you help you number 1, 2 nad 3

SJCC Instructor: Lazik Chemistry 1A Practice Exam 3 This worksheet is intended for students enrolled in SJCC CHEM 1A (Dr. lyun Lazik) and canno reproduced or distributed without explicit consent from this instructor t be Which of the following reactionés) will proceed to the product side? (a) MgCl2 (aq) + Br2 (l) MgBr (aq) + Cl2 (g) (b) 2 H20 (I) + 2K (s) 2 KOH (aq) + H2 (g) (c) Ba (s) + MgO (s)-> Bao (s) + Mg (s) 1. 2. Consider the formation of ammonia from combination of nitrogen gas and hydrogen gas, if AH for this reaction is-92 kJ, how much heat is released when 350 kg of nitrogen reacts with excess hydrogen gas? Draw an enthalpy diagram showing the enthalpy change for the formation of ammonia. Given the following equation and ah, determine the heat of formation of Chu (1). 2 GH14 (1) + 100 (g)- 12 CO2 (g) + 14 H20 (g) 3. alf-8326 ki

Explanation / Answer

1) ANSWER: b, C,

higher SRP having element/ion involves in reduction, lower one involves in oxidation.

   Br2 = srp = +1.066 v

   Cl2 = srp = +1.36 v

reaction a do not takes place.

b) K is a alkali metal reacts with water liberates H2 gas.

c) higher SRP having element/ion involves in reduction, lower one involves in oxidation.

it is takes place.

Ba , SRP = -2.912 V

Mg , SRP = -2.372 V

2)
        N2 + 3H2 ---> 2NH3   DH = -92 kj

   no of mol of N2 reacted = 350*10^3/17 = 2.06*10^4 mol

amount of heat released = 2.06*10^4*92/2 = 9.48*10^5 kj

3) from equation

   DH0rxn = (14*DH0f,H2O + 12*DH0f,CO2) - (2*DH0f,C6H14+19*DH0f,O2)

   -8326 = (14*-241.85 + 12*-393.5)-(2*X+19*0)

X = DH0f,C6H14 = +109.05 Kj/mol

ENTHALPY OF FORMATION OF C6H14 = +109.05 Kj/mol

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