Chapter 5

Questions :

1.Define these terms: system, surroundings, thermal energy, chemical energy, potential energy, kinetic energy, law of conservation of energy.

System : a specific part of the universe that is of interest to us .

Surrounding : the rest of the universe outside the system .

Thermal energy : the energy associated with the random motion of atoms and molecules .

Chemical energy : the energy stored within the structural units of a chemical substance .

Potential energy :the energy possessed by an object virtue to its position .

Kinetic energy : the energy results from motion .

 Law of conservation of energy : energy cannot be destroyed nor created , but it can change into another forms .

2.What is heat? How does heat differ from thermal energy? Under what condition is heat transferred from one system to another?

- Heat is the transfer of thermal energy  between two bodies,whereas thermal energy is the energy associated by the random motion of atoms and molecules .
- Heat is transferred from one system to another when there is a different in temperature . 

3.What are the units for energy commonly employed in chemistry?

The SI unit of energy is Joule , also there is  another unit for energy called cal 

$$1cal\,=\,4,184\,joule$$

4.A truck initially traveling at 60 km/h is brought to a complete stop at a traffic light. Does this change violate the law of conservation of energy? Explain.

No, this doesn't violate the law of conservation of energy because the kinetic energy is converted into heat through the friction.

5.Define these terms: thermochemistry, exothermic process, endothermic process.

Thermochemistry is the study of heat in chemical reaction.

6.Stoichiometry is based on the law of conservation of mass. On what law is thermochemistry based? 

It is based on the law of conservation energy.

7.Describe the interconversions of forms of energy occurring in these processes: (a) You throw a softball up into the air and catch it. (b) You switch on a flashlight. (c) You ride the ski lift to the top of the hill and then ski down. (d) You strike a match and let it burn completely

a) kinetic energy into potential energy into kinetic energy into potential energy again .

b) chemical energy into electrical energy .

c) potential energy into kinetic energy .

d) Thermal energy into chemical energy .

8.Decomposition reactions are usually endothermic, whereas combination reactions are usually exothermic. Give a qualitative explanation for these trends.

Energy is needed to break chemical bonds while energy is released when chemical bonds are formed .

9.On what law is the first law of thermodynamics based? Explain the sign conventions in the equation
$$\Delta U = q+w$$  
The fist law of thermodynamic is based on the law of conservation of energy .
when the sign of the previous equation is negative that means the energy is released , and if it is positive that means , by contrast , the energy is absorbed .

10.The work done to compress a gas is 47 J. As a result, 93 J of heat is given off to the surroundings. Calculate the change in internal energy of the gas 

$$\Delta U = q+w$$  
$$q=-93J$$
$$w=47J$$
then
$$\Delta U=-93+47=-46J$$

11.In a gas expansion, 87 J of heat is released to the surroundings and the energy of the system decreases by 128 J. Calculate the work done

$$\Delta U = q+w\,\rightarrow$$  
$$w=\Delta U-q$$
$$\Delta U=-128J$$
$$q=-87J$$
$$w=-128-(-87)=-41J$$

12.Calculate w, and determine whether work is done by the system or on the system when 415 J of heat is released and ΔU = 510 J.

$$\Delta U = q+w\,\rightarrow$$  
$$w=\Delta U-q$$
$$\Delta U=510J$$
$$q=-415J$$
$$w=510-(-415)=925J$$
$$The\,work\,is\,done\,on\,the\,system$$

13.Calculate q, and determine whether heat is absorbed or released when a system does work on the surroundings equal to 64 J and ΔU = 213 J

$$\Delta U = q+w\,\rightarrow$$  
$$q=\Delta U-w$$
$$\Delta U=213J$$
$$w=64J$$
$$q=213-(64)=149J$$
$$The\,heat\,is\,absorbed\,by\,the\,system$$

14.The diagram on the left shows a system before a process. Determine which of the diagrams on the right could represent the system after it undergoes a process in which (a) the system absorbs heat and ΔU is negative; (b) the system absorbs heat and does work on the surroundings; (c) the system releases heat and does work on the surroundings.

a) (ii)
b) (ii)
c) (ii)

15.The diagram , upward , on the left shows a system before a process. Determine which of the diagrams on the right could represent the system after it undergoes a process in which (a) work is done on the system and ΔU is negative; (b) the system releases heat and ΔU is positive; (f) the system absorbs heat and ΔU is positive

a) (iii)
b) (iii)
c) (iii)

16. Consider these changes.

(a) Hg(l) → Hg(g)

(b) 3O2(g) → 2O3(g)

(c) CuSO4 · 5H2O(s) → CuSO4(s) + 5H2O(g)

(d) H2(g) + F2(g) → 2HF(g)

At constant pressure, in which of the reactions is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done?

a) The work is done by the system .
b) The work is done on the system . " because fewer moles on the right mean that the gas is compressed .
c) The work is done by the system .
d) No work is done . "because the state didn't change "

17.Define these terms: enthalpy and enthalpy of reaction. Under what condition is the heat of a reaction equal to the enthalpy change of the same reaction?

-Enthalpy is the heat exchanged between the system and the surrounding .
-Enthalpy of reaction is the heat exchanged at a constant pressure for a specific reaction .
The enthalpy change is equal to the heat of reaction when the reaction is undergone at a constant volume .

18.In writing thermochemical equations, why is it important to indicate the physical state (i.e., gaseous, liquid, solid, or aqueous) of each substance?

because it helps to determine the actual enthalpy changes .

19.Consider this reaction:
$$2(CH_3OH)_{(l)}\,+3(O_2)_{(g)}\rightarrow 4(H_2O)_{(l)}+2(CO_2)_{(g)}$$
$$\Delta H =1452,8KJ/mole$$
What is the value of ΔH if (a) the equation is multiplied throughout by 2; (b) the direction of the reaction is reversed so that the products become the reactants, and vice versa; (c) water vapor instead of liquid water is formed as the product?

a) ΔH=2905.6 KJ/mole
b) ΔH=-1452.8KJ/mole
c) ΔH will be more than it is when the liquid water is formed because it will need more energy .

20.A sample of nitrogen gas expands in volume from 1.6 to 5.4 L at constant temperature. Calculate the work done in joules if the gas expands (a) against a vacuum, (b) against a constant pressure of 0.80 atm, and (c) against a constant pressure of 3.7 atm. (1 L · atm = 101.3 J.)

a) there is no work done against the vacuum .
b)
$$w=-P \Delta V$$
then : 
P=0.80 atm
and
$$\Delta V= (5.4-1.6=3.8L)$$
the work done is :
$$w=0.80 \times 3.8=-3.04 L.atm$$
converting from l.atm to Joule : $$-3.04 L.atm\times \frac{101.3 Joule}{1L.atm}=-307.952 joule$$

c)
$$w=-P \Delta V$$
then : 
P=3.7 atm
and
$$\Delta V= (5.4-1.6=3.8L)$$
the work done is :
$$w=-3.7 \times 3.8=-14.06 L.atm$$
converting from L.atm to Joule : $$-14.06 L.atm\times \frac{101.3 Joule}{1L.atm}=-1,424 Kj$$

21.A gas expands in volume from 26.7 to 89.3 mL at constant temperature. Calculate the work done (in joules) if the gas expands (a) against a vacuum, (b) against a constant pressure of 1.5 atm, and (c) against a constant pressure of 2.8 atm. (1 L · atm = 101.3 J.)

a) there is no work since the work is done against the vacuum .

b) 
$$w=-P \Delta V$$
then : 
P=1.5 atm
and
$$\Delta V= (89.3-26.7=62.2 mL)$$
the work done is :
$$w=-1.5 \times 0.0622L=-0.0933 L.atm$$
converting from L.atm to Joule :

$$-0.0933 L.atm\times \frac{101.3 Joule}{1L.atm}=-9.45joule$$

c) 
$$w=-P \Delta V$$
then : 
P=2.8 atm
and
$$\Delta V= (89.3-26.7=62.2 mL)$$
the work done is :
$$w=-2.8 \times 0.0622L=0.174 L.atm$$
converting from l.atm to Joule :

$$-0.174 L.atm\times \frac{101.3 Joule}{1L.atm}=-17.64joule$$

22.A gas expands and does PV work on the surroundings equal to 325 J. At the same time, it absorbs 127 J of heat from the surroundings. Calculate the change in energy of the gas.

$$\Delta U = q+ P \Delta V$$
$$P \Delta V = -325 J$$
$$q=127 J$$
then :
$$\Delta U = 127-325=-198J$$

23.The first step in the industrial recovery of zinc from the zinc sulfide ore is roasting, that is, the conversion of ZnS to ZnO by heating:

$$2ZnS+3O_2\rightarrow 2ZnO+2SO_2\; \Delta H=-879Kj/mole$$

To calculate the heat evolved of ZnS first step is to divide the heat of reaction by 2 then divide it by molar mass 65.41+32.07=97.48g/mole

$$\frac{-879KJ}{2mole}\times \frac{1mole}{97.48g}=4.51KJ/g$$

24.consider the following reaction :
$$2H_2O\; \rightarrow 2H+2O \Delta H =483.6KJ/mole$$
at a certain temperature. If the increase in volume is 32.7 L against an external pressure of 1.00 atm, calculate ΔU for this reaction. (1 L · atm = 101.3 J.)

$$\Delta U=q+w$$
$$w=P\Delta V=32.7 \times 1atm=32.7L.atm \times 101.3=3312.51 Joule$$
$$q=483600Joule/mole$$
$$\Delta U = 483600+3312.51=4.81\times10^2Joule$$

25. consider the following reaction :
$$H_2+Cl_2\rightarrow 2HCl \Delta H = -184.6 joule/mole$$
If 3 moles of H2 react with 3 moles of Cl2 to form HCl, calculate the work done (in joules) against a pressure of 1.0 atm. What is ΔU for this reaction? Assume the reaction goes to completion and that ΔV = 0. (1 L · atm = 101.3 J.)

the work done in this reaction is Zero since V is Zero .

$$\Delta U = q$$
$$q=3\Delta H = 3 \times -184.6 = -553.8Joule$$
"q is tripled because there are three moles of reactant are involved"
$$\Delta U = -553.8 Joule$$


26.The diagrams represent systems before and after reaction for two related chemical processes. ΔH for the first reaction is −595.8 kJ/mol. Determine the value of ΔH for the second reaction.

$$\Delta H = +595.8$$
The change in enthalpy is the same in the magnitude and opposite in sign because the first reaction is combustion and the second one is decomposition .

27. For most biological processes, the changes in internal energy are approximately equal to the changes in enthalpy. Explain.

Because these kind of reaction are taken place in a constant volume so there is no work done .

28.What is the difference between specific heat and heat capacity? What are the units for these two quantities? Which is the intensive property and which is the extensive property?


Main difference  The difference is that the specific heat is for the heat needed to raise up one gram of a substance one C temperature , but the heat capacity is the heat needed to raise up a sample of substance by one C temperature .
Units :
Heat capacity : 
$$\frac{Joule}{C^o}$$
Specific Heat :
$$\frac{Joule}{g.C^o}$$
Extensive or Intensive :
Heat capacity is Extensive but Specific heat is Intensive .

29.Define calorimetry and describe two commonly used calorimeters . In a calorimetric measurement , why is it important to know the heat capacity of the calorimeter ? How is this value determined ?

Calorimetry is the study of heat changes 

There are two known calorimeter :
1. crude constant pressure calorimeter .
2. constant volume bomb .

It is important to specify the heat capacity of the calorimeter because it is important in measurement of "q" [heat change between surrounding and system]


This table may be helpful in solving some problems :

30. A 6.22-Kg piece of copper metal is heated from 20.5C of 324.3C. calculate the heat absorbed (in kJ) by the metal .

$$q=sm\Delta T$$
$$s_{cu}=0.385J/g.C$$
$$m=6.22\times10^3 g$$
$$\Delta T =324.3C-20.5C=303.8C$$
$$q=0.385 \times 6,22 \times 10^3 \times 303.8=727.509 KJ$$

31.Calculate the amount of heat liberated (in kJ) from 366 g of mercury when it cools from 77.0°C to 12.0°C

$$q=sm\Delta T$$
$$s_{Hg}=0.139 J/g.C$$
$$m=366 g$$
$$\Delta T=77-12=-65$$
$$q=0.139\times 366 \times -65 =-3.307Kj$$

32.A sheet of gold weighing 10.0 g and at a temperature of 18.0°C is placed flat on a sheet of iron weighing 20.0 g and at a temperature of 55.6°C. What is the final temperature of the combined metals? Assume that no heat is lost to the surroundings. (Hint: The heat gained by the gold must be equal to the heat lost by the iron.

$$q_{Au}=-q_{Fe}$$
$$q_{Au}=0.129 \times 10 \times (\Delta T_{final}-18C)$$
$$q_{Au}=1.29\Delta T - 23.22 J$$
$$q_{Fe}=0.444\times 20 \times (\Delta T_{final}-55.6C)$$
$$q_{Fe}=8.88\Delta T - 493.728 J$$

33.A 0.1375-g sample of solid magnesium is burned in a constant-volume bomb calorimeter that has a heat capacity of 3024 J/°C. The temperature increases by 1.126°C. Calculate the heat given off by the burning Mg, in kJ/g and in kJ/mol.

the heat gained by calorimeter is
$$q=C\Delta T$$
$$3024 \times 1.126=3405.024 J$$
The heat given off by Mg per gram is
$$3405.024 J \div 0.1375 g = 24.764 KJ/g$$
The heat given off by Mg per mole is :
$$\frac{24.765}{1g} \times \frac{24.31g}{1mol}=602Kj/mole$$

34.A quantity of 2.00 × 10^2 mL of 0.862 M HCl is mixed with 2.00 × 10^2 mL of 0.431 M Ba(OH)_2 in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the HCl and Ba(OH)2 solutions is the same at 20.48°C. For the process:
$$H^++OH^-=H_2O$$
the heat of neutralization is −56.2 kJ/mol. What is the final temperature of the mixed solution? Assume the specific heat of the solution is the same as that for pure water.

$$q_{sln}=-q_{rxn}$$
There is 0.1724(=2 × 0.0862) mole of HCl involved in the reaction and 0.0862 mole of Ba(OH)_2
that means the number of moles of reactant is stoichiometrically equivalent according to the equation :
$$2HCl + Ba(OH)_{2}=2H_2O+BaCl_2$$
$$q_{rxn}=-56.2 \times 10^3 J$$
The amount of heat released when 1 mole of H 'positive'
is neutralized is given in the problem (-56.2 kJ/mol). The 
amount of heat liberated when 0.172 mole of H'positive'
is neutralized is:
$$0.1724 mole \times \frac{-56.2 \times 10^3J}{1mole}$$  
q of reaction=9.69\times 10^3 J
$$q_{sln}=sm\Delta T$$
by assuming the specific heat of the reaction is the specific heat of the water and the density of the solution is the density of water 
$$q_{sln}=4.184\times 400 times (\Delta T _{final}-20.48)$$

q of solution =1673.6 × Delta T - 34275

673.6  T{final} - 34275=9.69 × 10^3 

$$T _{final}=\frac{43965}{1673.6}=26.2 C$$

35. A 50.75-g sample of water at 75.6°C is added to a sample of water at 24.1°C in a constant-pressure calorimeter. If the final temperature of the combined water is 39.4°C and the heat capacity of the calorimeter is 26.3 J/°C, calculate the mass of the water originally in the calorimeter.

$$q_{sln}=-q_{rxn}$$
q of solution is the heat capacity times the difference in temperature plus the heat absorbed by initially placed water .
$$q_{sln}=(26.3\times 15.3)+(4.148\times m \times 15.3)$$
$$q_{sln}=402.4 + 63.5 j$$
$$q_{rxn}=sm\times \Delta T$$
$$q_{rxn}=4.148 \times 50.75 \times (-36.2)$$
$$q_{rxn}=-7620.5$$
$$-7620.5=-402.4-63.5 m$$
$$m=\frac{7218.1}{63.5}=113.7 g$$

36.A piece of silver with a mass of 362 g has a heat capacity of 85.7 J/°C. What is the specific heat of silver?

$$s=C/g$$
$$s=\frac{85.7}{362}=0.236(J/g.C^o)$$

37.Consider two metals A and B, each having a mass of 100 g and an initial temperature of 20°C. The specific heat of A is larger than that of B. Under the same heating conditions, which metal would take longer to reach a temperature of 21°C?

Metal A will take longer time than metal B because it has larger specific heat .


38. From these data :

calculate the enthalpy change for the transformation :

$$S_{rhombic}\rightarrow S_{monoclinic}$$
(Monoclinic and rhombic are different allotropic forms of elemental sulfur.)

The enthalpy change can be illustrated according to Hess's law as the following :
$$S_{rhombic}+O_2\rightarrow SO_2\; \Delta H =-296.06KJ/mole$$
$$SO_2\rightarrow  \; S_{monoclinic}+O_2\Delta H =296.36 KJ/mole$$
By applying Hess's law The enthalpy change is :
$$\Delta H = -296.06 Kj+296.36 Kj=300 Joule$$

39. From the following data :

calculate the enthalpy change of 

$$2C_{graphite}+3H_2 \rightarrow C_2H_6$$

the calculation of enthalpy change can be illustrated according to Hess's law as the following :

$$2C+2O_2\rightarrow 2CO_2$$

$$3H_2+\frac{3}{2}O_2\rightarrow 3H_2O$$ $$2CO_2+3H_2O\rightarrow C_2H_6+\frac{7}{2}O_2$$

Then 

$$\Delta H =$$