Question

A mixture of 0.5540 M CO and 0.3180 M Cl2 is enclosed in a vessel and heated to 1000 K. cog) +Cl2(g) 근 COC12(g) Kc =255.0 at 1000 K Calculate the equilibrium concentration of each gas at 1000 K. Number Number Number

Answer 1

Here, Cl_2 is the limiting reagent, since, 0.3180 M < 0.5540M CO (g) + Cl_2 (g) rightharponoverleftharpon CoCl_2 (g) Initially: 0.5540 M 0.3180 0 At eqlm: 0.554 - 0.318 alpha 0.318 (1 - alpha) 0.318 alpha where alpha = degree of disappearance of Cl_2 So, K_c = [CoCl_2]_eq/[Co]_eq [Cl_2]_eq = 0.318 alpha/(0.554 - 0.318 alpha) 0.318 (1 - alpha) rightdoublearrow 255 = alpha/(0.554 - 0.318 alpha) (1 - alpha) rightdoublearrow 255 = alpha/0.554 - 0.872 alpha + 0.318 alpha^2 rightdoublearrow 141.27 - 222.36 alpha + 81.09 alpha^2 = alpha rightdoublearrow 81.09 alpha^2 - 223.36 alpha + 141.27 = 0 rightdoublearrow alpha = 223.36 plusminus squareroot (- 223.36)^2 - 4 times 81.09 times 141.27/2 times 81.09 = 223.36 plusminus 63.77579792/162.18 rightdoublearrow alpha = 223.36 - 63.77579792/162.18 = 0.98399434 so, [CO]_eq = (0.554 - 0.318 times 0.98399434) M = 0.241089799 M [Cl_2]_eq = 0.318 times (1 - 0.98399434) M = 0.00508979988 M [COCl_2]_eq = 0.318 times 0.98399434 M = 0.3129102 M