Question

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At a certain temperature the equilibrium constant, Kc, equals 0.11 for the reaction: 2 IC1(g)=12(8) + Cl2(g). What is the equilibrium concentration of Cl2 if 3.18 mol of I2 and 3.18 mol of Cl2 are initially mixed in a 2.0-L flask?

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Answer 1

kindly post different questions separately.

concentration (M) = mole Yolumell) Volume of flouk = 2.0L initial concentration of Cl2 = 3.18 mol = 1.59 M 2.0L initial concentration of I2 = 3.18 mol = 1.59M 2.0L goes in As initally no amount of ICL is present, so the reaction revene direction to attain equilibrium ICE Table concentration (M) 2 tolg) I219) + Cl 19) 1159 1.59 tax -x -x E 2x 1159 -30 1. 59-30 At equilibrium CED kes equilibrium constant = 0ill - Ke: [12] [Cl ] = Oill = (1.59-30) (1159-x) (ICIJ2 (20) 2 LOJT 1:59-30 Taking square root on both side, We get value of x and a canot 0.33) = 1.59 - 2 have negative value or greater than 1.59 ) 0.3377 X 2x = 1:59 - = 0.6634 x + x = 1.59. x = 0.956 M At equilibrium, concentration of Cla [612] = 1.59-x = (1,59 -0.956) M | [12] = 0.634 M Answer con