10. Using data tabulated below ( Page 5), calculate ° at 25°C for NAs) + 02(g) 2

Question

10. Using data tabulated below ( Page 5), calculate ° at 25°C for NAs) + 02(g) 2 NC)(g). a. -8655 kJ mol-1 b 86.55 kJ mol-1 d. This cannot be determined without additional information c. 173.10 kJ mol-1 e. none of these. 11. Given /..--504 kJ-mor, and as, --372 J-K.,mol1. calculate the value of Graat 358 K and determine in J-K mol.calculate the value of AG, at 38 K and determine in which direction the reaction is spontaneous at 358 K for the reaction below: to, (s) + 4HRg) + F, (g) UF, (s) + 2H2O(g) A) AG 371 kJ-mol': spontaneous in the forward direction B) AG.. 371 kJ mol; spontaneous in the reverse direction C) 0," _ 637 kJ-mol-1, spontaneous in the reverse direction D) Gr.--637 k mori, spontaneous in the forward direction E) None of these is true 12. Arrange the following reaction equations according to increasing AS values i. C.H, (g) + 30, (g) 2CO2 (g) + 2H,O (g) ii. PCI s (g) PCI, (g) + Cl2(g) iii. Na (s) +0, (g) Nad), (s) iv. 2 C.Hio (g) + 13 02 (g) 8 CO, (g) + 10 H,O (g) a. i

Explanation / Answer

10)

Gorxn = Gfo(products) - Gfo( reactants)

= 2Gfo [NO(g)] - Gfo [N2(g)] - Gfo [O2(g)]

= 2 x 86.55 kJ/mol - 0 - 0

= 173.1 kJ/mol

Therefore,

Gorxn = 173.1 kJ/mol

11)

Given that

H= -504 kJ/mol = - 504000 J/mol
S= - 372 J/K/mol

T = 358 K

For any reaction to be spontaneous, G < 0

The relation between G, H, S is

G = H - TS

     = - 504000 J/mol - [ 358 K x - 372 J/K/mol ]

     = - 370824 J/mol

     = -371 kJ/mol

      < 0

Therefore,

the reaction is spontaneous in forward direction.

12) Ans: C

    order of entropy : gas > liquid > solid

   In eq (iii) , s + g -------> s

   Therefore, it should have least entropy.

   Therefore, ans= c

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