Question

4. Classify each of the following reactions as one of these four types:

• spontaneous at all temperatures

• not spontaneous at any temperature

• spontaneous below a certain temperature but not above

• spontaneous above a certain temperature but not below

(a) CO (g) + 3 H2 (g) → CH4 (g) + H2O (g); ∆H = -206.1 kJ; ∆S = -214.6 J/K

(b) AgClO3 (s) + CH4 (g) → AgCl (s) + 2 H2O (g) + CO (g); ∆H = -616 kJ; ∆S = 343 J/K

(c) CaS (s) + H2S (g) → CaH2 (s) + 2 S (s); ∆H = 317 kJ; ∆S = -157 J/K

(d) 2 NH3 (g) → N2 (g) + 3 H2 (g); ∆H = 92.2 kJ; ∆S = 198.8 J/K

Answer 1

Spontaneity of a reaction can be evaluated by calculating the change in free energy of the reaction using the relation:

$\Delta$G = $\Delta$ H - T$\Delta$S

for a spontaneous reaction $\Delta$ G <0

thus temperature for the reaction to be spontaneous, $\Delta$ H - T$\Delta$S <0

or T>$\Delta$H /$\Delta$S

a) CO (g) + 3 H2 (g) → CH4 (g) + H2O (g); ∆H = -206.1 kJ; ∆S = -214.6 J/K

Given that $\Delta$ H = -206.1 kJ and $\Delta$ S = -214.6 J/K = -0.2146 kJ/K

$\Delta$G = -206.1 - T*(-0.2146)

$\Delta$G = -206.1 + T*0.2146

thus $\Delta$ G will be < 0 for 206.1>T*0.2146

or T< 960 K

Thus the reaction will be spontaneous for temperature below 960 K

(b) AgClO3 (s) + CH4 (g) → AgCl (s) + 2 H2O (g) + CO (g); ∆H = -616 kJ; ∆S = 343 J/K

Given that $\Delta$ H = -616 kJ and $\Delta$ S = 343 J/K = 0.343 kJ/K

$\Delta$G = -616 - T*(0.343)

$\Delta$G = -206.1 - T*0.343

since T> 0

thus $\Delta$ G will be < 0 for all values of T

Therefore reaction is spontaneous at all temperatures

(c) CaS (s) + H2S (g) → CaH2 (s) + 2 S (s); ∆H = 317 kJ; ∆S = -157 J/K

Given that $\Delta$ H =317 kJ and $\Delta$ S = -157 J/K = -0.157 kJ/K

$\Delta$G = 317 - T*(-0.157)

$\Delta$G = 317 + T*0.157

as T > 0. thus $\Delta$ G >0 at all temperatures

Thus the reaction will be non-spontaneous at all temperatures.

(d) 2 NH3 (g) → N2 (g) + 3 H2 (g); ∆H = 92.2 kJ; ∆S = 198.8 J/K

Given that $\Delta$ H =92.2 kJ and $\Delta$ S = 198.8 J/K = 0.1988 kJ/K

$\Delta$G = 92.2 - T*(0.1988)

$\Delta$G = 92.2 - T*0.1988

thus $\Delta$ G will be < 0 for 92.2<T*0.157

or T> 587 K

Thus the reaction will be spontaneous for temperature above 587 K