Question

For a given reaction, ∆H = 35.5 kJ mol-1 and ∆S = 83.6JK-mol-1. The reaction is spontaneous at : (Assume that ∆H and ∆S do not vary with temperature)
(1) T< 425 K
(2) T > 425 K
(3) all temperatures
(4) T > 298 K​

Answer 1

AnSwer :-

According to Gibbs-Helmholtz equation,

∆G = ∆H - T∆S

here

• ∆G = Gibbs energy
• ∆H = Enthalpy change
• ∆S = Entropy change
• T = temperature

For a reaction to be spontaneous ∆G < 0

$\therefore$ Gibbs-Helmholtz equation becomes,

∆G = ∆H – T∆S < 0 (or) ∆H < T∆S

$\sf T< \dfrac{∆H}{∆S} = \dfrac{35.5KJ {mol}^{ - 1} }{83.6J {K}^{ - 1} {mol}^{ - 1} }$

$\sf = \dfrac{35.5 \times 1000}{83.6} = 425K$

$\sf \therefore T > 425K$

Hence, option (2) T > 425K is correct.