Define third law of thermodynamics. Give its molecular interpret

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 Multiple Choice QuestionsShort Answer Type

291. Give at least two statements of second law of thermodynamics.
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 Multiple Choice QuestionsLong Answer Type

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292. Define third law of thermodynamics. Give its molecular interpretation. What is the most important application of this law?


Third law of thermodynamics states: 'The entropy of all perfectly crystalline solids may be taken as zero at absolute zero.'

Molecular interpretation: Entropy is a measure of disorder. Thus, at absolute zero a perfectly crystalline solid has a perfect order of its constituent particles i.e. there is no disorder at all. Hence absolute entropy is taken as zero.
Application. It helps in the calculation of the absolute entropies of the substances at room temperature (or at any temperature T). A simplified expression for the absolute entropy of solids at temperature T is given by,



where Cp is the heat capacity of the substance at constant pressure.

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293.

Explain the effect of temperature on feasibility for:

(i) endothermic process

(ii) exothermic process in terms of Gibb’s Helmoltz equation.

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 Multiple Choice QuestionsShort Answer Type

294.

When does endothermic processs become spontaneous? Explain.

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295. How is the concept of coupling of reactions useful in explaining the occurrence of non-spontaneous reaction?
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 Multiple Choice QuestionsLong Answer Type

296.

(a) What are Exergonic and Endergonic reactions?

(b) Given that the standard heat of formation of NH3(g) as represented by the equation

1 half straight N subscript 2 left parenthesis straight g right parenthesis space plus space 3 over 2 straight H subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space NH subscript 3 left parenthesis straight g right parenthesis

is – 46.191 kJ. The standard entropies of N2(g), H2(g) and NH3(g) are 191.62, 130.12 and 193.3 JK–1 mol–1 respectively. Calculate the standard free energy of formation (∆G0) for NH3. Is the reaction feasible?

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297.

(i) Define first and the second law of thermodynamics in the combined form. 
(ii)  
(a) Write an expression for the entropy change of an ideal gas for isothermal change.

(b) Write an expression for entropy change for an ideal gas for isobaric change.

(iii) Calculate the maximum work obtained when 0.75 ml of an ideal gas expands isothermally and reversibly at 27°C from a volume of 15L to 25L.

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298.

(i) What are the limitations of criterion for randomness?
(ii) Calculate the standard free energy of formation of straight H subscript 2 straight O subscript 2.The free change for the reaction.
straight H subscript 2 straight O subscript 2 space space rightwards arrow space space space straight H subscript 2 straight O space plus space 1 half straight O subscript 2 space space is
space space space space space space space space increment straight G to the power of 0 space equals space minus 125.10 space kJ space and
increment straight G subscript straight f superscript 0 left parenthesis straight H subscript 2 straight O right parenthesis space equals space minus 228.4 space kJ space mol to the power of negative 1 end exponent

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 Multiple Choice QuestionsShort Answer Type

299.

Given:
space straight N subscript 2 left parenthesis straight g right parenthesis space plus space space 3 straight H subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space 2 NH subscript 3 left parenthesis straight g right parenthesis semicolon space space increment subscript straight r straight H to the power of 0 space equals space minus 92.4 space kJ space mol to the power of negative 1 end exponent
What is the standard enthalpy of formation of NH subscript 3 gas?


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 Multiple Choice QuestionsLong Answer Type

300.

Calculate the enthalpy change for the process
         CCl subscript 4 left parenthesis straight g right parenthesis space space space rightwards arrow space space straight C left parenthesis straight g right parenthesis space plus space 4 space Cl left parenthesis straight g right parenthesis
and calculate bond enthalpy of straight C minus Cl in CCl subscript 4 left parenthesis straight g right parenthesis.
increment subscript vap straight H to the power of 0 space left parenthesis CCl subscript 4 right parenthesis space equals space 30.5 space kJ space mol to the power of negative 1 end exponent
increment subscript straight f straight H to the power of 0 space left parenthesis CCl subscript 4 right parenthesis space equals space minus 135.5 space kJ space mol to the power of negative 1 end exponent
increment subscript straight a straight H to the power of 0 left parenthesis straight C right parenthesis space equals space 715.0 space kJ space mol to the power of negative 1 end exponent comma space where space increment subscript straight a straight H to the power of 0 space is space enthalpy space of space atomisation
increment subscript straight a straight H to the power of 0 left parenthesis Cl subscript 2 right parenthesis space equals space 242 space kJ space mol to the power of negative 1 end exponent

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