Question

For a given reaction, $\Delta H$ = 35.5 kJ mol⁻¹ and DS = 83.6 JK⁻¹ mol⁻¹. The reaction is spontaneous at : (Assume that $\Delta H$ and $\Delta S$ do not vary with tempearature)
(a) T > 425 K
(b) All temperatures
(c) T > 298 K
(d) T < 425 K

Answer 1

Explanation:

check the solution in picture

Answer 2

Answer: (a) T > 425 K

Explanation:

According to Gibb's equation:

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

$\Delta G$ = Gibbs free energy  

$\Delta H$ = enthalpy change  = 35.5 kJ/mol = 35500 J/mol

$\Delta S$ = entropy change  =  83.6 J/Kmol

T = temperature in Kelvin

$\Delta G$= +ve, reaction is non spontaneous

$\Delta G$= -ve, reaction is spontaneous

$\Delta G$= 0, reaction is in equilibrium

$\Delta G=(+ve)-T(+ve)$

$\Delta G=(+ve)(-ve)=-ve$  

when $T\times \Delta S$> $\Delta H$

Thus $T\times 83.6J/Kmol>35500J/mol$

$T>425Kl$

Thus reaction is spontaneous at temperatures greater than 425 Kelvin