Question

A student determines the value of the equilibrium constant to be 1.13x10 for the following reaction. 3Fe2O3(s) + H2(9)—2Fe3O4(s) + H20(9) Based on this value of Keg: AGº for this reaction is expected to be greater, less) than zero. Calculate the free energy change for the reaction of 1.52 moles of Fe,O3(s) at standard conditions at 298K. AGⓇxn =

Answer 1

Solution ) we know that - AG= - RT en keq where = 8.3145 J/Kmal Tis Temperature in Kelvin keq is equil 'brium constant. Given Te 298k Keq = 1.13x10° Here (keq> 1/ so value of es keq will be positive so overall value of AG it will be negative. negative value are always less than zero. AG less than zero Ans. Gow= - RT en keq = -(0.3145 J/xmae) (290 Klen(1.13X105) = -34533. 2018 3 J/mol Grer = -34.53 kJ /mol Anse above solution For given reaction - 35e03 (5) + H2(g) 2 Fe3 04 (s) + H2O(g)

( 3High corg) For Chh (g) + H2O(g) Given keq = 2.34 X10-25 we know that AG = - RT enkeq Here keq LL1 So en keq is a negative value. gives a negative and overall value of AG will be positive A positive value is always greater than zero Tog is greater than zero Ans. - Gon = Ten keq Doon = -(2.3145] Ie'me') (2986) ((2,34x1025) are = (–2477,721 I mal-) (-56.7144764) O Grown = 140522.6492 J | modi Gions = 140.52 KJ | mad ADS.