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

Step-by-step explanation:

Formula used :

`\Delta G^o=-2.303* RT* \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* 8.314* 298* \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