Thermodynamics

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Payal Gupta

Contributor-Level 10

23. (a) The state of a thermodynamic system is described by its measurable or macroscopic (bulk) properties. We can describe the state of a gas by quoting its pressure (p), volume (V), temperature (T), amount (n) etc. Variables like p, V, T are called state variables or state functions because their values depend only on the state of the system and not on how it is reached. Example: Volume of water in a pond, is a state function, because change in volume of its water is independent of the route by which water is filled in the pond, either by rain or by tube well or by both.

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Payal Gupta

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22. qrev = (-ΔfH? ) = - (- 286kJ mol-1) = 286 x 103 Jmol-1 = 286000 J mol-1

ΔS  (surroundings) qrev/ T = 286000 J mol-1 / 298 K

                        =  959.73 J mol-1 K-1

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Payal Gupta

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21. For NO (g), ΔrH° is a +ve value. So, it is unstable in nature.

For NO (g), ΔrH° is a -ve value. So, it is stable

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Payal Gupta

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20. ΔG° = -RT ln K = - 2.303 RT log K

Putting the values of

R = 8.314 J K-1 mol-1,

T = 300 K and

K =10; we get

ΔG° = - 2.303 x 8.314 J K-1 mol-1 x 300 K and K x log 10

        = - 5527 J mol-1

        = -5.527 kJ mol-1

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Payal Gupta

Contributor-Level 10

19. ΔH° = ΔU° + Δng RT

ΔU° = -10.5 kJ, Δng = 2-3 = -1 mol, R = 8.314 x 10-3 kJ mol-1, T = 298 K

ΔH° = (- 10.5 kJ) + [ (- 1 mol) x (8.314 x 10-3 kJ mol-1) x (298 K)]

       = -10.5 kJ – 2.478 kJ

       = -12.978 kJ

According to Gibbs Helmholtz equation:

ΔG° = ΔH° - TΔS°

        = (- 12.978 kJ) – (298 K) x (- 0.0441 kJ K-1)

        = -12.978 + 13.142

        = 0.164 kJ

Since the value of ΔG° is positive, the reaction is non-spontaneous.

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Payal Gupta

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18. ? H: negative (-ve) because energy is released in bond formation

? S: negative (-ve) because entropy decreases when atoms combine to form molecules.

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Payal Gupta

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17. As per the Gibbs Helmholtz's equation:

ΔG = Δ H - TΔ S

For ΔG=0; 

ΔH=TΔS

Or  T=ΔH/ΔS

T = (400 KJ mol-1)/ (0.2 KJ K-1 mol-1)

= 2000 k

Thus, reaction will be in a state of equilibrium at 2000 K and will be spontaneous above this temperature

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Payal Gupta

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16. Change in internal energy (? U) for an isolated system is zero for it does not exchange any energy with the surroundings. But entropy tends to increase in case of spontaneous reaction. Therefore? S > 0 or positive.

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Payal Gupta

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15. According to the question:

i.  CCl4  (l) → CCl4  (g),                        ? vapH? = 30.5 kJ mol–1

iii. C (s) + 2Cl2 (g)  CCl4  (l),              ? fH? = –135.5 kJ mol–1

iii. C (s) → C (g),                                  ? aH? = 715.0 kJ mol–1

iv Cl2  (g) → C (g) + 4 Cl (g)                 ? aH? = 242 kJ mol–1

The equation we want is:

CCl

...more

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Payal Gupta

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14. The reaction for the formation of CH3OH can be obtained by:

C (s) + 2H2 (g) + l/2O2 (g) → CH3OH (l)

This can be obtained by the algebraic calculations of the reactions:

Equation (ii) + 2 x equation (iii) – equation (i)

? Hf  [CH3OH] =? cH? + 2 x? fH? -? rH?

                        = (– 393 kJ mol-1) + 2 (- 286 kJ mol-1) – (– 726 kJ mol-1)

                        = (– 393 – 572 + 726) kJ

...more

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