Is entropy of universe constant? ///Second law of thermodynamics.
Class 11 Chemistry
Is entropy of universe constant?
Answer:- No, the entropy of the universe always increases in the course of every spontaneous change. It is the second law of thermodynamics.
Define combined form of first and second law of thermodynamics.
Answer:- In thermodynamics, the combined law of thermodynamics, also called the Gibb’s fundamental equation, is a mathematical summation of the first law of thermodynamics and the second law of thermodynamics summed into a single concise mathematical statement as
dU- TdS + PdV ≤ 0
Where dU is a variation in internal energy.
T is temperature, dS is variation in entropy
P is pressure
dV is variation in volume of a simple working body in which there is neither flows of particles out of the body nor external forces, other than gravity, acting on the body. In theoretical structure, in addition to the obvious inclusion of the first two laws, the combined law incorporates the implications of the zeroth law, via temperature T and the third law, through its use of free energy as related to the calculation of chemical affinities near absolute zero.
What will be the value of internal energy for one mole of an ideal gas ?
Answer:
Internal energy (Eint) of the ideal gas is defined as the,
Eint = 3/2 nRT
Where n is the number of moles, R is the gas constant and T is the temperature.
To find out the internal energy of the ideal gas, substitute 1 mole for n,
250° C for temperature T and 8.31 J/ mol K for the gas constant R in the equation.
E = 3/2 nRT
= 3/2 (1 mol) (8.31 J/ mol. K) (250°C)
= (1.5)( 1 mol) (8.31 J/ mol. K) (250 + 273) K
= (1.5)( 1 mol) (8.31 J/ mol. K) (523) K
= 6519.195 J or 6520 J
Energy is neither absorbed nor released during expansion of ideal gas in vacuum. Why?
Answer:- In an ideal gas, there are no intermolecular forces of attraction. Hence, no energy is required to overcome these forces. Like, when a gas expends against vacuum, work done is zero as Pext – Hence, internal energy of the system does not change, there is no absorption nor release of energy during expansion of ideal gas in vacuum.
When will bond energy be equal to bond dissociation energy?
Answer:- When bond energy is measured in isolated gaseous state, bond energy becomes equal to bond dissociation energy.
Why enthalpy changes whereas, internal energy does not change due to change in heat energy ?
Answer:- Mostly chemical reaction occurs under constant atmospheric pressure. In this situation, heat change occurs on the system is different from the change under constant volume.
Hence, heat energy change occurs under constant pressure but internal energy does not change due to this heat energy change but enthalpy changes.
How non-spontaneous process be converted to spontaneous process?
Answer:- AG = ∆H + TAS; when the signs for ∆H and ∆S are both the same, then temperature determines that the process is spontaneous. When ∆H is positive (unfavourable) and ∆S is positive (favourable), high temperatures are needed so the favourable ∆S term dominates making the process spontaneous (∆G < 0). When ∆H is negative (favourable) and ∆S is negative (unfavourable), low temperatures are needed so the favourable ∆H term dominates making the process spontaneous (∆G < 0).
Differentiate between isothermal and adiabatic process?
Answer:- Process in which temperature remains constant throughout the process is called isothermal process. When such a process occurs, heat can flow from the system to the surrounding and vice versa in order to keep the temperature constant. In an isothermal process, dT= 0.
When a process is carried out in such a way that no heat can flow from the system to the surroundings or vice versa i.e., system is completely insulated from the surroundings. In such a system, temperature of the system always changes. For an adiabatic process dQ = 0
Internal energy change is state function but work is not a state function. Why?
Answer:- Internal energy is state function because it describes quantitatively an equilibrium state of a thermodynamic system, irrespective of how the system arrived in that state. In contrast, work is process quantity or path function, because its value depends on the specific transition (or path) between two equilibrium states.
Which of the two, diamond or graphite, has more enthalpy?
Answer:- Graphite has greater enthalpy since it is loosely packed.
What is the relation between E and H?
Answer:-
E = Internal energy, H = Enthalpy,
∆H = ∆E + P∆V or ∆H = ∆E + ∆ngRT
here ∆E = Change in internal energy
∆H = Enthalpy change,
∆ ng = Number of moles of gaseous products – Number of moles of geseous reactants Where R = Gas constant
T = Temperature
In which state a substance has maximum entropy?
Answer:- Entropy by definition is the degree of randomness in a system. Out of three states of matter: Solid, Liquid and Gas, the gaseous particles move freely and therefore, the degree of randomness is the highest. Solids have highly ordered arrangement and has low entropy. The entropy of liquids is more than solids because molecules of liquid shows less ordered arrangement than solids. The molecules of gas are free to move and not constrained to be adjacent to each other. Therefore, Sgas >> Sliquid > Ssolid. In gaseous state, entropy is maximum.
Give examples of endothermic and exothermic reactions.
Answer:- Reaction in which heat is given out along with the products is called exothermic reaction.
Example:
1. Burning of methane gas releases a large amount of energy. Hence it is an exothermic reaction.
2. Reaction of calcium oxide and water
Reactions that absorb energy or require energy in order to proceed are called endothermic reactions.
Example :
(a) When ammonium chloride is dissolved in water in a test tube, the test tube becomes cold.
(b) Reaction of barium hydroxide with ammonium chloride.
What is the heat of neutralisation of NH4 OH and HCl?
Answer:- Enthalpy of neutralization of ammonium hydroxide (weak base) and hydrochloric acid (strong acid) is -51.5 kJ.
Write Gibb’s Helmholtz Equation.
Answer:-
Gibb’s Helmholtz Equation can be given as :
∆G = ∆H – T∆S
Where; ∆G = Free energy change
∆H = Enthalpy change
∆S = Entropy change
T = Absolute temperature
What will be the efficiency of Carnot engine if its source is at 500 K and sink at 300 K?
Answer:
Efficiency, η = (T2-T1)/T2
T2 = 500 K (temperature of the source)
T1 = 300 K (temperature of the sink)
Therefore,
η = (T2 – T2])/r2
= (500 – 300) / 500
= 200/500 .
= 2/5
= 0.4
What are the necessary conditions for adiabatic process?
Answer:
The necessary conditions for adiabatic process are—
(a) There should not be any exchange of heat between the system and its surrounding. All walls of the container and the piston must be perfectly insulating.
(b) The system should be compressed or allowed to expand suddenly so that there is no time for the exchange of heat between the system and its surroundings.
Question 23.
Heat of neutralisation of strong acid and strong base is constant. Why?
Answer:
The enthalpy of neutralization of all strong acids and strong bases is always constant, i.e., -57.1 kJ. The explanation to this generalization can be provided on the basis of Arrhenius theory of ionization. The strong acids and strong bases are almost completely ionized in dilute aqueous solutions. The neutralization of strong acid and strong bases simply involve the combination of H+ ions (from acid) and OH– ions (from base) to form water molecules. It has been found experimentally that when 1 mole of water is formed by the neutralization of 1 mole of H+ ions and 1 mol of OH– in aqueous solutions, 57.1 kJ of energy is released.
Question 24.
Though dissolution of NaCl in water is an endothermic reaction, it is soluble in water. Explain.
Answer:
The enthalpy of solution of NaCl in water (that is, the energy change associated with the dissolution of sodium chloride crystals in water) at standard conditions is very slightly positive, it is an endothermic process. A positive enthalpy change indicates that the system acquires heat from the surroundings in order for the reaction (dissolution) to proceed forward. When something dissolves in water, some of these O-H bonds are broken. This requires heat energy. Entropy increases when solute dissolves in a solvent. Also, if H is positive, G will be negative and process is spontaneous.
an example of isolated system.
Answer:
Thermos flask is the best example of isolated system. A thermos flask is used to keep things either cold or hot. Thus, a thermos does not allow energy for transfer. Additionally, the thermos, like any other closed container, does not allow matter transfer because it has a lid that does not allow anything to enter or leave the container. An isolated system does not exchange energy or matter with its surroundings.
In which process of system, temperature decreases?
Answer:
Adiabatic process is the process in which change in pressure, volume and temperature takes place without any heat entering or leaving the system is called adiabatic process. So the total heat of the system, undergoing an adiabatic process always remains constant. In an adiabatic expansion since no heat is supplied from outside, therefore the energy required for the expansion of the gas is taken from the gas itself. This signifies that, the internal energy of an ideal gas undergoing in an adiabatic expansion decreases, and because the internal energy of an ideal gas depends only on the temperature, therefore its temperature must decreases. That is why the temperature of a gas drops in an adiabatic expansion.
Explain second law of thermodynamics with the help of efficiency of Carnot engine. How free energy is a measure of spontaneity of a reaction?
Answer:
According to second law of thermodynamics, :
“All spontaneous processes are thermodynamically irreversible”.
Or
“Without the help of an external agency, a spontaneous process cannot be reversed”.
The change of heat into mechanical work is done by Carnot heat engine. In real engines, it is not possible to convert heat energy completely into work because heat is lost in friction, as radiation etc. Hence, Carnot calculated efficiency and reversible work by ideal engine.
In this engine, at standard conditions, source of heat (at T2K) and after work, to store rest of heat,a sink (at T2K) is required. As a source of heat, working substance, ideal gas is taken in such a cylinder whose walls are insulators but bottom is conducting. A frictionless piston is required in insulators. The Carnot cycle when acting as a heat engine consists of the following four steps: two isothermal and two adiabatic processes. Four stages are A, B, C and D, In whole Carnot cycle, at temperature T1, q1 heat is absorbed from the source and w work is done and rest part q2 is given in the sink at temperature T2. The work done by Carnot engine is given by the following formula,
w = q1 – q2 = q1(T2 -T1)/T1
In this way, in a cyclic process, heat absorbed by the source cannot be converted to equivalent amount of work without any loss of heat. The efficiency of Carnot Engine can be calculated by this. “That absorbed heat which can be converted to work is called efficiency of Carnot engine.
n = w/q1 = (T2 -T1)/T1.
(T2 – T1)/T1 < 1
Hence, efficiency of Carnot engine is always less than one. This proves second law of thermodynamics, according to it, “heat can neither be converted to equivalent amount of work”.
The efficiency of Carnot engine explains the entropy concept of second law of thermodynamics.
Free energy (∆G) gives a criteria for spontaneity at constant pressure and temperature,
If ∆G is negative (<0), the process is spontaneous.
If ∆G is positive (>0), the process is non spontaneous.
Question 39.
Explain the following:
1. Enthalpy of formation
2. Phase Transition enthalpy
3. Entropy
4. Enthalpy of solution
Answer:
(1) Enthalpy of Formation : It is a form of standard enthalpy of reaction in which one mole of compound is formed from its elements. The standard enthalpy change for the formation of one mole of a compound from its elements. The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation is called Standard Molar Enthalpy of Formation. Its symbol is ∆H°.
N2(g) + O2(g) ➝ 2NO(l) ; ∆fH° =+180 kJ
(ii) Phase Transition Enthalpy: The change in phase of a substance takes place at a very small rate. Thus, it is not possible to measure the heat change for phase transition experimentally. But, phase transition enthalpy can be calculated with the help of Hess’s Law. The heat of transition of diamond to graphite can be calculated from the heat of combustion data. For diamond and graphite, which is -395.4 kJ and -393.5 kJ respectively.
(3) Entropy : Entropy is a measure of the molecular disorder, or randomness, of a system. The concept of entropy provides deep insight into the direction of spontaneous change for many everyday phenomena. Its introduction by the German physicist Rudolf Clausius in 1850 is a highlight of 19th-century physics.
There is increase in entropy during melting of a solid and vaporisation of a liquid, in solids, there is a definite crystal lattice whereas in gases and liquids, this arrangement is very less. Therefore, with increase in entropy in gases and liquids, disorderlness also increases. Thus, entropy increases in spontaneous processes. Therefore, entropy of a system is a measure of orderlness. In this way, entropy changes related to molecules take place.
- More the number of molecules, more will be the entropy.
- More the number of energy levels available to molecules on heating, more will be the entropy.
4. Enthalpy of solution : Enthalpy of solution of a substance is the enthalpy change when one mole of it dissolves in a specified amount of solvent. The enthalpy of solution at infinite dilution is the enthalpy change observed on dissolving the substance in an infinite amount of solvent when the interactions between the ions (or solute molecules) are negligible.
When an ionic compound dissolves in a solvent, the ions leave their ordered positions on the crystal lattice. These are now more free in solution. But solvation of these ions (hydration in case solvent is water) also occurs at the same time. Thus, in dissolution of ionic solid, contributing of both the energies is necessary.
Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol-1 K-1.
Answer:
The expression of heat (q) is written as follows :
q = m.c. ∆T ….(i)
Where, c = molar heat capacity = 24 J mol-1 K-1
m = mass of substance = 60 g = 60/27 mol
∆T = change in temperature = 20°C (35°C to 55°C)
Substituting the values in equation (i), we get
q = m.c.∆T
q = (60/27) × 24 × 20
= 1066.7 J q
= 1.07 kJ
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