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Thermodynamics

Short Answer Type

291. Give at least two statements of second law of thermodynamics.

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Long Answer Type

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

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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.

Gibb’s Helmoltz equation is

(i) Endothermic process: ∆H is positive, thus it always opposes the process. Now,

(a) If T∆S is negative and opposes the process, then ∆G will be positive and the process is always non-spontaneous.

(b) When T∆S is positive i.e. favourable, then ∆G may either be positive or negative.

At low temperature: T∆S may have a small value and ∆H may be greater than T∆S. Under these conditions, ∆G(=∆ H + T∆S) may be positive and the reaction may not be spontaneous at low temperature.

Increasing temperature: With the increase in temperature, the magnitude of the favourable factor T∆S increases while ∆H does not change much. Hence at high temperature, the magnitude of T∆S will be quite large and more than ∆H so that ∆G becomes negative. This means that endothermic processes are favoured and are more probable at high temperature.

(ii) Exothermic process: ∆H is always negative and, therefore, it is favourable. Now,

(a) If T∆S is positive i.e. favourable then ∆G have only negative value and the process is spontaneous at all temperatures.

(b) If T∆S is negative i.e. unfavourable, then ∆G can have positive or negative value.

At high temperature: –T∆S will have large magnitude and ∆G will be positive and as such, the process may not be spontaneous.
At low temperature. The value of ∆H may become greater than the small value of T∆S and ∆G becomes negative and the process is spontaneous under these conditions. Hence exothermic processes are favoured and more probable at low temperature.

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Short 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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Long 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?

114 Views

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.
$straight H subscript 2 straight O subscript 2 space space rightwards arrow space space space straight H subscript 2 straight O space plus space 1 half straight O subscript 2 space space is space space space space space space space space increment straight G to the power of 0 space equals space minus 125.10 space kJ space and increment straight G subscript straight f superscript 0 left parenthesis straight H subscript 2 straight O right parenthesis space equals space minus 228.4 space kJ space mol to the power of negative 1 end exponent$

144 Views

Short 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?

124 Views

Long 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 $CCl subscript 4 left parenthesis straight g right parenthesis.$
$increment subscript vap straight H to the power of 0 space left parenthesis CCl subscript 4 right parenthesis space equals space 30.5 space kJ space mol to the power of negative 1 end exponent increment subscript straight f straight H to the power of 0 space left parenthesis CCl subscript 4 right parenthesis space equals space minus 135.5 space kJ space mol to the power of negative 1 end exponent increment subscript straight a straight H to the power of 0 left parenthesis straight C right parenthesis space equals space 715.0 space kJ space mol to the power of negative 1 end exponent comma space where space increment subscript straight a straight H to the power of 0 space is space enthalpy space of space atomisation increment subscript straight a straight H to the power of 0 left parenthesis Cl subscript 2 right parenthesis space equals space 242 space kJ space mol to the power of negative 1 end exponent$