Thermochemistry And Electrochemistry Problems

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1.The standard heats of formation at 298K for CCl4(g),H2O(g),CO2(g) and HCl(g) are -25.5,-57.8,-94.1 and -22.1 kcal/mole respectively.Calculate the ΔH0 298for this reaction.
Answer:-41.4 kcal

2.How much heat will be required to make 1kg of calcium carbide according to the following reaction:
CaO(s) + 3C(s) -----------CaC2(s) + CO(g)
The heat of formation og CaO,CaC2 and CO are -151.6,-14.2 and -26.4 kcal respectively.
Answer:1732 kcal

3.When one mole of ethyl alcohol is completely burnt in oxygen to form water and carbon dioxide,327kcals are evolved at 25°C.Calculate the heat of reaction at constant volume.
Answer:-326.4 kcal

4.Calculate the difference between heats of reaction at constant pressure and constant volume for the reaction at 25°C in kJ.
C6H6 (l) + 15O2(g)-------- 12CO2(g) +6H2O (l)
Answer:-7.4 kJ

5.Calculate the heat of formation of acetic acid from the following data:
(a) C(s) + O2(g)-------- CO2(g) ΔH=-94kcal
(b) H2(g) + 1/2 O2(g)------------H2O(l) ΔH= -68kcal
...
... (c) CH3COOH(l) + 2O2(g)-------2CO2(g) + 2H2O(l) ΔH=-208kcal
Answer:-116kcal

6.Calculate the heat of the following reaction:
CH2=CH2(g) + H2(g)-------CH3--CH3(g)
The bond energy of C-C,C=C,C-H and H-H bonds are 80kcal,145kcal,98kcal and 103 kcal respectively.
Answer: -28kcal

7.Thermochemical reactions
C(gra) + 1/2 O2(g)----- CO(g) ΔH=-110.5 kcal
CO(g) + 1/2O2(g)------CO2 (g) ΔH= -283.2 kcal
From the above reaction,the heat of reaction for C(graphite) + O2(g)---CO2(g) will be?
Answer: -393.7 kJ

8.Given:
C + 2S---- CS2 , ΔH0f = +117.0 kJ/mol
C + O2---- CO2, ΔH0f =-393 kJ/mol
S + O2---- SO2, ΔH0f =-297 kJ/mol
The heat of combustion of
CS2 + 3O2------- CO2+ 2SO2 ?
Answer: -1104 kJ/mol

9.ΔH0f of CO2(g),CO(g),N2O(g) and NO2(g) in kJ/mol re respectively -393. -110, +81 and +34 respectively.Calculate ΔH in kJ of the following reaction:
2N02(g) + 2CO(g)---- N2O + 3CO2(g)
Answer: -836 kJ

10.H2 + 1/2 O2------ H 2O ; ΔH= -68.39 kcal
K + H2O----KOH + 1/2 H2 ; ΔH = -48 kcal
KOH + water------KOH(aq); ΔH= -14 kcal
The heat of formation of KOH(in kcal) is.....
Answer:+26.3 kcal

Electro chemistry deals with chemical changes produced by passing electricity.The laws which govern the decomposition of substances on electrodes with passage of current are called Faraday's laws of electrolysis

PROBLEMS ON FARADAY'S I LAW ( electrochemistry problems)

1. Calculate the amount of sodium formed at the cathode by passing 965coulombs of electricity through molten sodium chloride.
Solution: NaCl---------Na+ +Cl-

At cathode:Na+ + e - ----------Na

23 grams of sodium is deposited by passing 96500 coulombs of electricity.

------- grams of sodium will be deposited by passing 965 coulombs of electricity.

965 x 23/ 96500 =.23grams

Check this Activation Energy Formula awesome i recently used to see.

The amount of sodium formed by passing 965 coulombs of electricity is .23grams.
2. Calculate the amount of products formed at the electrodes when 2 Faraday's of electricity is passed through i)molten NaCl ii)aqueous NaCl.

Solution

i) When electricity is passed through sodium chloride the following reactions take place

NaCl---------Na+ +Cl-

At Cathode:Na+ +e- ---------Na

23grams of sodium is deposited by passing 1Faraday of electricity or 96,500coulombs f electricity.

-------- grams of sodium is deposited by passing 2Faraday's of electricity

23x2=46grams

At Anode:2Cl - ---------Cl2 +2e-

35.5grams(equivalent weight of chlorine)is deposited by passing 1Faraday of electricity.

------grams will be deposited by passing 2Farady's of electricity.

35.5x2=71grams.

ii) when aqueous sodium chloride is subjected to electrolysis the following reactions take place

NaCl---------Na+ +Cl-

H2O--------H+ +OH-

At cathode2:H+ +2e- ------------H2

Na+ do not get discharged at the cathode in the presence of H+ as they require more potential.

1gram of hydrogen gets deposited by passing 1faraday of electricity.

-----grams will be deposited by passing 2faradays of electricity.

1x2=2grams of hydrogen.

At Anode:2Cl - ---------Cl2 +2e- (chloride ions require lesser potential to get discharged compared to hydroxide ions so they get deposited at anode in preference to hydroxide ions)

35.5grams(equivalent weight of chlorine)is deposited by passing 1Faraday of electricity.

------grams will be deposited by passing 2Farady's of electricity.

35.5x2=71grams.
3. Calculate the time required for deposition of 4grams of Magnesium when 365 ampere of current is passed through fused magnesium chloride

solution: MgCl2-----------Mg+2 +2Cl-

cathode: Mg+2 + 2e---------------------- Mg

24 grams of magnesium is deposited by passing 2 Faraday of electricity (2x96,500c)

4 grams is deposited by 25733.3C

C = 365 amperes

Q = C x t

25733.3 = 365 x t

t = 25733.3/365

t = 70.50

FARADAY'S SECOND LAW (electrochemistry problems)

QUESTION: An electric current was passed through two voltameter cells containing copper sulphate (using copper electrodes)and the other containing silver nitrate solution (using silver electrodes) .The increase in weight of cathodes in two cases was respectively 0.189gm and 0.648 gm .Calculate the chemical equivalent of copper, taking that of silver as 108.

SOLUTION:

Weight of copper deposited = 0.189 gm

Weight of silver deposited = 0.648 gm

Eq. wt of silver = 108

Eq. wt of copper = ?

According to Faraday's second law

Wt of copper deposited = Eq.wt of copper

Wt . of silver deposited Eq.wt of silver

0.189/0.648 = Eq.wt of copper/108

= 0.189x108/0.648

Eq.wt of copper = 31.5

Learn more on about Modern Periodic Table and its Examples. Between, if you have problem on these topics louis de broglie atomic theory, Please share your comments.