Answer: The equilibrium constant for the reaction at 25 °C is 346.7
Explanation:
Formula used :
[tex]\Delta G^o=-2.303\times RT\times \log K_c[/tex]
where,
[tex]\Delta G^o[/tex] = standard Gibb's free energy change = -14.50kJ/mol =14500 J/mol
R = universal gas constant = 8.314 J/K/mole
T = temperature = [tex]25^0C= (25+273)K=298 K[/tex]
[tex]K_c[/tex] = equilibrium constant = ?
Putting in the values we get:
[tex]-14500=-2.303\times 8.314\times 298\times \log K_c[/tex]
[tex]\log K_c=2.54[/tex]
[tex]K_c=antilog(2.54)=346.7[/tex]
The equilibrium constant for the reaction at 25 °C is 346.7
