Question

Tedict and explain whether the follo ind explain whether the following reactions will be nonspontaneous at all temperatures or spontaneous at all, low, or high temperatures. (a) 2POCI(8) ► 2PC(s) + O2(8) Air - 572 3; AS - 179 3K (b) 302(g) → 203(g) AF - 286 kJ; AS = -1373/ (c) 4NH3(g) + 502(g) → 4NO(g) + 6H2O() AH = -1274 kJ; AS = 180 J/K (d) 2PbS(s) + 302(g) → 2PbO(s) + 2502(8) AH = -844 kJ; AS = -165 3/K a (e) The reaction referenced in Question 1 26 DESSA Wasmo 51655 POSNEBUDERUS

Answer 1

a)

ΔHo = 572.0 KJ

ΔSo = 179 J/K

= 0.179 KJ/K

use:

ΔGo = ΔHo - T*ΔSo

for reaction to be spontaneous, ΔGo should be negative

that is ΔGo<0

since ΔGo = ΔHo - T*ΔSo

so, ΔHo - T*ΔSo < 0

572.0- T *0.179 < 0

T *0.179 > 572.0

T > 3196 K

Answer: at high temperatures

b)

ΔHo = 286.0 KJ

ΔSo = -137 J/K

= -0.137 KJ/K

use:

ΔGo = ΔHo - T*ΔSo

for reaction to be spontaneous, ΔGo should be negative

that is ΔGo<0

since ΔGo = ΔHo - T*ΔSo

so, ΔHo - T*ΔSo < 0

286.0- T *-0.137 < 0

T *0.137 < -286.0

T < -2087.5912 K

Since kelvin temperature can't be negative

The reaction will never be spontaneous

Answer: at no temperatures

c)

ΔHo = -1274.0 KJ

ΔSo = 180 J/K

= 0.18 KJ/K

use:

ΔGo = ΔHo - T*ΔSo

for reaction to be spontaneous, ΔGo should be negative

that is ΔGo<0

since ΔGo = ΔHo - T*ΔSo

so, ΔHo - T*ΔSo < 0

-1274.0- T *0.18 < 0

T *0.18 > -1274.0

T > -7077.7778 K

Since kelvin temperature can't be negative

The reaction is spontaneous at all temperatures

The reaction will always be spontaneous

Answer: at all temperatures

d)

ΔHo = -844.0 KJ

ΔSo = -165 J/K

= -0.165 KJ/K

use:

ΔGo = ΔHo - T*ΔSo

for reaction to be spontaneous, ΔGo should be negative

that is ΔGo<0

since ΔGo = ΔHo - T*ΔSo

so, ΔHo - T*ΔSo < 0

-844.0- T *-0.165 < 0

T *0.165 < 844.0

T < 5115 K

Answer: at low temperatures

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