For a given reaction, ΔH = +35.5 kJ/mol and ΔS = +83.6 J/Kmol. The reaction is spontaneous . Assume that ΔH and ΔS do not vary with temperature.

a) at T > 298 K
b) at T > 425 K
c) at all temperatures
d) at T < 425 K

b) at T > 425 K

Explanation:

ΔG = ΔH - TΔS

∴ ΔG < 0 ⇒ The reaction is spontaneous

∴ ΔG > 0 ⇒ The reaction is not spontaneous

∴ ΔG = 0 ⇒ The equilibrium

∴ ΔH = +35.5 KJ/mol

∴ ΔS = (+83.6 J/K.mol)*(KJ/1000 J) = 0.0836 KJ/K.mol

at T = 298 K:

⇒ ΔG = 35.5 KJ/mol

⇒ ΔG = 35.5 - (298)(0.0836) = 35.5 - 24.913 = 10.587 KJ/mol > 0... No spontaneous

at T = 425 K:

⇒ ΔG = 35.5 - (425)(0.0836) = 35.5 - 35.53 = - 0.03 KJ/mol < 0is spontaneous

The reaction is spontaneous at temperature: T > 425. Choice B.

The mathematical representation of the Gibb's free energy as given below provides a basis for determining if a reaction is spontaneous or not.

ΔG = ΔH - TΔS ΔG < 0 ⇒ The reaction is spontaneous ΔG > 0 ⇒ The reaction is not spontaneous ΔG = 0 ⇒ The equation is in equilibrium

The value of the enthalpy and entropy given are;

ΔH = +35.5 KJ/molΔS = (+83.6 J/K.mol)*(KJ/1000 J) = 0.0836 KJ/K.mol

By setting;

ΔG < 0

Therefore;

35.5 - T × (0.0836) < 035.5 < 0.0836T

By dividing through by 0.0836; we have;

35.5/0.0836 < T425 < T

In essence, the reaction is spontaneous at temperature, T > 425 in which case the Gibb's free energy: ΔG < 0.

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