Ke & Equilibrium Pressures and in • The equil. constant, kp, for the following rxn is 10.5 @ 350k 2 CH₂Cl2 cg) - Chucgs + CCl4 cgi If an equil. mixture of the 3 gases in a 13 L container @ 350 K contains CH₂Cl2 e a pressure of 0.558 atm . CHu o a pressure of 0.272 atm, the equil. PP of coly is atm.

L Le Châtelier Calculations : [ ] The equil. constant, K, for the following rxn is 1.80 x 102 @ 698 K. 2 H11g - H₂ (g) + 12 (g) An equil. mixture of the 3 gases in all flask @ 698 K contains 0.329 M HI, 4.41 x 102 M H₂ , & 4.41 x 102 M 12. What will be the [ ] of the 3 gases once equil has been reestablished, if 2.41 x 102 mol of H₂ ig). is added to the flask?

Calculate K: From initial 1 Final [] A student ran the following ran in the lab @ 289 Ki 1 2 CH₂ Cl2 (g) = CHUCg & CCl4 con When she introduced 66.71 x 10-2 mol of CH₂Cl2 (g) into a 1 L container, she found the equil. [] of CC14 (9) to be 3.11 x 10-2 M. Calculate the equil. constant, ke, she obtained for this ran. Kc = ?

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Fur the giua reaction Kp as Kp lo5 p4 o 272 atm p eg ce 0358 atm on puttin alu... View the full answer