The equilibrium constant, K. for the following reaction is 2.77E-2 at 623 K: COC

Question

The equilibrium constant, K. for the following reaction is 2.77E-2 at 623 K: COCl2(g) CO(g) + CL2(g) An equilibrium mixture of the three gases in a 11.2 L container at 623 K contains 0.246 M COCI2, 8.26E-2 M CO and 8.26E-2 M CI2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 4.49 L? [COCI2] = M. [CO] = M. [Cl2] = M.

Explanation / Answer

COCl2 ----> CO + Cl2 0.246-x 0.0826+x 0.0826+x Kc = [Cl2][CO]/[COCl2] 0.027 = (0.082+x)^2/11.2x(0.246-x) x = 0.1158 at equi [COCl2] = 0.246-0.1158 = 0.1302/4.9 = 0.02657 [CO] =( 0.0826+0.1158)/4.9 = 0.0405 [Cl2] = 0.0405

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