Determine whether the reactions with the following ΔH and ΔS values

1.	Determine whether the reactions with the following ΔH and ΔS values are spontaneous or non-spontaneous. State whether they are exothermic or endothermic.(a) ΔH = -110 kJ and ΔS = +40 JK-1 at 400 K(b) ΔH = +50 kJ and ΔS = -130 JK-1 at 250 K.
Answer» (a) Given : ΔH = -110 kJ ΔS = 40 JK^-1 = 0.04 kJK Temperature = T = 400 K ΔG = ? Since ΔH is negative, The reaction is exothermic. ΔG = ΔH – TΔS = -110 – 400 × 0.04 = -110 – 16 = -126 kJ Since ΔG is negative, The reaction is spontaneous. (b) Given : ΔH = 50 kJ, ΔS = -130 JK^-1 = -0.13 kJ K^-1 Temperature = T = 250 K ΔG = ? Since ΔH is positive, The reaction is endothermic. ΔG = ΔH – TΔS = 50 – 250 × (-0.13) = 50 + 32.5 = 82.5 kJ Since ΔG > 0, The reaction is non-spontaneous. ∴ (a) ΔG = -126 kJ; The reaction is exothermic and spontaneous. (b) ΔG = 82.5 kJ; The reaction is endothermic and non-spontaneous.

Determine whether the reactions with the following ΔH and ΔS values are spontaneous or non-spontaneous. State whether they are exothermic or endothermic.(a) ΔH = -110 kJ and ΔS = +40 JK-1 at 400 K(b) ΔH = +50 kJ and ΔS = -130 JK-1 at 250 K.

Answer»

(a) Given :

ΔH = -110 kJ

ΔS = 40 JK^-1 = 0.04 kJK

Temperature = T = 400 K

ΔG = ?

Since ΔH is negative,

The reaction is exothermic.

ΔG = ΔH – TΔS

= -110 – 400 × 0.04

= -110 – 16

= -126 kJ

Since ΔG is negative,

The reaction is spontaneous.

(b) Given :

ΔH = 50 kJ,

ΔS = -130 JK^-1 = -0.13 kJ K^-1

Temperature = T = 250 K

ΔG = ?

Since ΔH is positive,

The reaction is endothermic.

ΔG = ΔH – TΔS

= 50 – 250 × (-0.13)

= 50 + 32.5

= 82.5 kJ

Since ΔG > 0,

The reaction is non-spontaneous.

∴ (a) ΔG = -126 kJ;

The reaction is exothermic and spontaneous.

(b) ΔG = 82.5 kJ;

The reaction is endothermic and non-spontaneous.

Discussion

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