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crimeas [40]
3 years ago
14

A reaction has a standard free-energy change of -14.50 kJ mol(-3.466 kcal mol). Calculate the equilibrium constant for the react

ion at 25 °C. 5.85 K
Chemistry
1 answer:
den301095 [7]3 years ago
5 0

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

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