Question

Question 18 Tries remaining 1 Points out of 100 p Pag question For the equilibrium reaction written below, find AG (kJ) at 298 K. The concentration of C2H2 is 0.26 M, the concentration of C6H6 is 0.081 M, and the standard change in Gibb's free energy is -111.9 kJ for this reaction. 3CH2(6) CH.Ca) Multiple tries are permitted; however, 20% (1/5) point will be deducted for each incorrect response.

Answer 1

standard change in free energy - Temperature gas constant We Knowe Relation between free energy Change , equilibrium constant & Reaction Quotient as AG - AG + RT ln Reaction Quotient We have given AG=-11-9 kJ R = 8.31 J K mot T = 298 K 3 Cg Hg (9) - Cho (9) a = [co Ho ] - 0.081 0.081 -4.6081 0.017576 [934973 How AG=-111.9+8.31 x 298 ln 5 =-111-9 + 2476.38X1G b on 5 = 1.6 Cappo) =-111.9 + 3962.208 = 3850. 308 kJ Hence for this reaction AG 18 3850 308 kJ.