Question

Anwar the following question​

Answer 1

Explanation:

4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)

H



f

[NO(g)] = 90.4

mol

kJ

H



f

[H2O(g)] = -241.8

mol

kJ

H



f

[NH3(g)] = -46.3

mol

kJ

H



rxn

= (4)(90.4

mol

kJ

) + (6)(-241.8

mol

kJ

) - [(4)(-46.3

mol

kJ

) + (5)(0)] = -904.0

mol

kJ

If you include moles as units for your coefficients: H



rxn

= -904.0 kJ

2. Calculate the entropy change for 2Na(s) + Cl2(g)  2NaCl(s)

S values: Na(s) = 51.05 J/mol•K, Cl2(g) = 223.0 J/mol•K, NaCl(s) = 72.38 J/mol•K

S



rxn

= (2)













K mol

J

72.38 - (2)













K mol

J

51.05 - (1)













K mol

J

223.0

= -180.3

K mol

J



If you include moles as units for your coefficients: S



rxn

= -180.3

K

J

3. a) Calculate Ggiven H = -227 kJ, S = -309 J/K, T = 1450 K.

G = H - TS



























 

 

J

k J

K

J

G k J K

1000

309 1

 227 (1450 ) 221 kJ

b) Is this process spontaneous at this temperature? If not, calculate the temperature (in oC) at

which this reaction becomes spontaneous.

No, it is not spontaneous at 1450 K (G is +)

Set G = 0: 0 = H - TS  T =

S

H





T =

































J

kJ

K

J

kJ

1000

1

309

227

= 735 K -273 = 462 o

C  T at equilibrium

H and S are both negative, so it is spontaneous at temperatures below 462 o

C.

Spontaneous at T < 462 oC