Class 12 chemistry chapter 4 Chemical Thermodynamics solutions

Maharashtra state Chemistry Textbook Solutions for Class 12 are very important and crusial that helps the students in understanding the complex topics and helps them in the preparation of class 12 board examination as well as verious compititive entrance examinations also. Studying the answers to the questions in the Chemistry textbook will check your understanding of a particular topic and helps you determine your strengths and weaknesses.

Class 12 chemistry textbook Solutions for Class 12, Chemistry  Chapter 4Chemical Thermodynamics maharashtra state board are provided here with simple step-by-step  detailed explanations. These solutions for  Chemical Thermodynamics very popular among Class 12 students for chemistry chapter 4 Chemical Thermodynamics Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the chemistry textbook Solutions Book of Class 12 chemistry Chapter 3 are provided here for you for free. You will also love the experience on ybstudy class 12 Solutions. All chemistry textbook Solutions. Solutions for class 12, These chemistry textbook solutions are prepared by  Chemistry experts and are 100% accurate

1. Select the most apropriate option.
i. The correct thermodynamic 
conditions for the spontaneous 
reaction at all temperatures are
a. ∆H < 0 and ∆S > 0
b. ∆H > 0 and ∆S < 0
c. ∆H < 0 and ∆S < 0
d. ∆H < 0 and ∆S = 0

ii. A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 bar from an initial volume of 2.5 L to a final volume of 4.5 L. The change in internal energy, ∆U of the gas will be
a. -500 J
b. + 500 J
c. -1013 J
d. + 1013 J

iii. In which of the following, entropy of 
the system decreases?
a. Crystallization of liquid into solid
b. Temperature of crystalline solid is
increased from 0 K to 115 K
c. H2(g) —-> 2H(g)
d. 2 NaHCO3(s) —-> Na2CO3
(s) + CO2(g) + H2O(g)

iv. The enthalpy of formation for all 
elements in their standard states is
a. unity
b. zero
c. less than zero
d. different elements

v. Which of the following reactions is 
a. H2(g) —-> 2H(g)
b. C(s) —–> C(g)
c. 2 Cl(g) —–> Cl2(g)
d. H2O(s) —–> H2O(l)

vi. 6.24 g of ethanol are vaporized by 
supplying 5.89 kJ of heat. Enthalpy 
of vaporization of ethanol will be
a. 43.4 kJ mol-1
b. 60.2 kJ mol-1
c. 38.9 kJ mol-1
d. 20.4 kJ mol-1

vii. If the standard enthalpy of formation of methanol is -238.9 kJ mol-1 then entropy change of the surroundings 
will be
a. -801.7 J K-1
b. 801.7 J K-1
c. 0.8017 J K-1
d. -0.8017 J K-1

viii. Which of the following are not state 
1. Q + W
2. Q
3. W
4. H-TS
a. 1,2 and 3
b. 2 and 3
c. 1 and 4
d. 2,3 and 4

ix. For vaporization of water at 1 bar, 
∆H = 40.63 kJ mol-1 and ∆S = 108.8 J K-1 mol-1. At what temperature, 
∆G = 0 ?
a. 273.4 K
b. 393.4 K
c. 373.4 K
d. 293.4 K

2. Answer the following in one or two sentences.
i. Comment on the statement: no work 
is involved in an expansion of gas in 
Answer : For a free expansion into a vacuum, a gas does no work, as there is no resistance on the gas as it expands, i.e. P=0 and therefore the quantity −PΔV=0

ii. State the first law of thermodynamics.
Answer : First law of thermodynamics is simply the conservation of energy. According to this law the total
energy of a system and surroundings remains constant when the system changes from an initial state to final state.
the first law of thermodynamics is ΔU = Q − W. Here ΔU is the change in internal energy U of the system.

iii. What is enthalpy of fusion?
Answer : Heat of fusion’ measures the amount of energy needed to melt a given mass of a solid at its melting point temperature. Conversely, it also represent the amount of energy given up when a given mass of liquid solidifies. Water, for example, has a heat of fusion of 80 calories per gram.

iv. What is standard state of a substance?
Answer : All chemical substances are either solid, liquid or gas. To make comparisons easier, the chemistry community has agreed on a concept called “the standard state.” The standard state of a chemical substance is its phase (solid, liquid, gas) at 25.0 °C and one atmosphere pressure.

v. State whether ∆S is positive, negative 
or zero for the reaction 2H(g) —–> 
H2(g). Explain.
Answer : 2h=H2 its wright delta s is possitive

vi. State second law of thermodynamics 
in terms of entropy.
Answer : The Second Law of Thermodynamics states that the state of entropy of the entire universe, as an isolated system, will always increase over time. The second law also states that the changes in the entropy in the universe can never be negative.

vii. If the enthalpy change of a reaction is ∆H how will you calculate entropy of surroundings?
Answer : The change in entropy of the surroundings after a chemical reaction at constant pressure and temperature can be expressed by the formula
ΔSsurr = -ΔH/T
ΔSsurr is the change in entropy of the surroundings
-ΔH is heat of reaction
T = Absolute Temperature in Kelvin.

viii. Comment on spontaneity of reactions for which ∆H is positive and ∆S is negative.
Answer : If ΔH is negative, and –TΔS positive, the reaction will be spontaneous at low temperatures (decreasing the magnitude of the entropy term). If ΔH is positive, and –TΔS negative, the reaction will be spontaneous at high temperatures (increasing the magnitude of the entropy term).

3. Answer in brief.
i. Obtain the relationship between ∆G°
of a reaction and the equilibrium 
Answer : A spontaneous reaction has a negative delta G and a large K value. … When delta G is equal to zero and K is around one, the reaction is at equilibrium. You have learned the relationship linking these two properties. This relationship allows us to relate the standard free energy change to the equilibrium constant.

ii. What is entropy? Give its units.
Answer :  Entropy is the measure of the disorder of a system. It is an extensive property of a thermodynamic system, which means its value changes depending on the amount of matter that is present. In equations, entropy is usually denoted by the letter S and has units of joules per kelvin (J⋅K−1) or kg⋅m2⋅s−2⋅K−1

iii. How will you calculate reaction enthalpy from data on bond enthalpies?
Answer :

iv. What is the standard enthalpy of combustion ? Give an example.
Answer : The standard enthalpy of combustion of a substnce is the standard enthalpy change accompanying a reaction in which one mole of the substance in its standard state is completely oxidised.

Consider the reaction
C2H2(g) + 5/2 x O2(g) ——> 2 CO2(g) + H2O(l), ∆r
H°= -1300 kJ

In the above reaction, the standard enthalpy change of the oxidation reaction, -1300 kJ is the standard enthalpy of combustion of C2H2(g).

v. What is the enthalpy of atomization? Give an example.
Answer : Enthalpy of atomisation, ΔaH0, is the change in enthalpy when one mole of bonds are completely broken to obtain atoms in the gas phase. For example: atomization of methane molecule. For diatomic molecules, enthalpy of atomization is equal to the enthalpy of bond dissociation.

vi. Obtain the expression for work done in chemical reaction.
Answer : The work done by a system at constant temperature and pressure is given by
W = Pext ∆V. Assuming Pext = P,
W = – P∆V
= – P (V2 – V1)
= – PV2 + PV1
II the gases were ideal, using Eq. (4.26)
PV1= n1RT and PV2
 = n2RT
At constant temperature and pressure.
W = – n2
RT + n1RT
= – (n2- n1) RT
= – ∆n
gRT (4.28)
The above equation gives the work done by
the system in chemical reactions. The sign of W depends on ∆V. We consider the following cases:
i. If n2
> n1, ∆ng
 is positive and W < 0

or work is done by the system.
ii. If n1
> n2, ∆ng
 is negative and W > 0
or work is done on the system.

iii. If n1= n2, ∆ng
= 0 and W = 0, or
No PV work is done when number of moles of reactants and products are equal.

vii. Derive the expression for PV work
Answer : Expression for pressure-volume (PV) work : Consider a certain amount of gas at constant pressure P is enclosed in a cylinder fitted with frictionless, rigid movable piston of area A.

 pressure is equal in magnitude and
opposite in sign to the external atmospheric pressure that opposes the movement and has its value -Pext. Thus,
f = -Pext × A (4.1)
where Pext is the external atmospheric pressure.

W = f × d (4.2)
Substitution from Eq. (4.1) gives
W = – Pext × A × d (4.3)
The product of area of the piston and
distance it moves is the volume change (∆V)
in the system.
∆V = A × d (4.4)
Combining equations (4.3) and (4.4) we write
W = – Pex ∆V (4.5)
 W = – Pex (V2 – V1)
where V2 is final volume of the gas.
When the gas expands, work is done by the
system on the surroundings. Since V2 > V1, W is negative. When the gas is compressed, work is done on the system by surroundings. In this
case V1< V2, and -Pext ∆V or W is positive.

viii. What are intensive properties? 
Explain why density is intensive 
Answer : An intensive property is a property of matter that does not change as the amount of matter changes. It is a bulk property, which means it is a physical property that is not dependent on the size or mass of a sample. In contrast, an extensive property is one that does depend on sample size.

Intensive properties are the properties which are independent of the mass or the extent of the system. Here, the density of the system remains unchanged even when it is divided into two parts. That means density is independent of the mass or the extent of the system. Thus, it is an intensive property.

Similar Posts

Leave a Reply

Your email address will not be published. Required fields are marked *