Question

3. 30 pts For a specific reaction: AG = 10.0 kJ/mole at 27.0°C AG = 2.00 kJ/mole at 177°C Assuming that the enthalpies and entropies of formation are independent of temperature for this system, calculate AS (reaction) and AH (reaction). Is this reaction favorable at all temperatures? Favorable at some temperatures? Never favorable? Explain your answer including key temperatures, if appropriate, in <25 words.

Answer 1

As ∆H and ∆S are independent of temperature, so for both 27 °C and 177 °C, the values will be same.

∆G = ∆H - T∆S

At 27 °C i.e. 27 + 273 = 300 K,

10 = ∆H - 300∆S. (Eqn 1)

At 177 °C i.e. 177 + 273 = 450 K,

2 = ∆H - 450∆S. (Eqn 2)

Subtratcting eqn 2 from eqn 1,

(10 - 2) = (∆H - ∆H) - (300∆S - 450∆S)

8 = 150∆S

∆S = 8/150 = + 0.0533 kJ/K•mol

∆S = + 53.3 J/K•mol

Substituting value of ∆S in eqn 1,

10 = ∆H - (300)(0.0533)

∆H = 10 + 16.00

∆H = + 26.00 KJ/mol

As given values of ∆G are negative, the reaction is not favorable at all temperatures. But ∆S being positive (favorable), ∆G can be negative at high temperature.

To find the temperature at which it is favorable:

Assume ∆G = 0

So, ∆H = T∆S

26.00 = T(0.0533)

T = 487.8 K

In °C, T = 487.8 - 273 = 214.8 °C

At T above 214.8 °C, ∆G would be negative and hence, reaction will be favorable.

To check, let the temperature = 215 °C

i.e. 215 + 273 = 488 K

∆G = 26.00 - (488)(0.0533)

= - 0.0104 kJ/mol

As ∆G is negative, reaction is favorable.

Hence, at temperature above 487.8 K (i.e. 214.8 °C) the reaction is favorable at all temperatures.