Question

A reaction has a standard free-energy change of -14.50 kJ mol(-3.466 kcal mol). Calculate the equilibrium constant for the reaction at 25 °C. 5.85 K

Answer 1

Answer:  The equilibrium constant for the reaction at 25 °C is 346.7

Explanation:

Formula used :

$\Delta G^o=-2.303\times RT\times \log K_c$

where,

$\Delta G^o$ = standard Gibb's free energy change = -14.50kJ/mol =14500 J/mol

R = universal gas constant = 8.314 J/K/mole

T = temperature = $25^0C= (25+273)K=298 K$

$K_c$ = equilibrium constant = ?

Putting in the values we get:

$-14500=-2.303\times 8.314\times 298\times \log K_c$

$\log K_c=2.54$

$K_c=antilog(2.54)=346.7$

The equilibrium constant for the reaction at 25 °C is 346.7