Be sure to answer all parts. The dissociation of molecular iodine into iodine atoms is represented as 1₂(g) = 21(g) At 1000 K, the equilibrium constant Ke for the reaction is 3.80 x 10-5. Suppose you start with 0.0461 mol of 1₂ in a 2.28-L flask at 1000 K. What are the concentrations of the gases at equilibrium? What is the equilibrium concentration of 1₂? M What is the equilibrium concentration of I? M

The equilibrium constant, K, for the following reaction is 6.59×10-2 at 546 K. PCl5(g) PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 6.59 L container at 546 K contains 0.254 M PCl5,   0.129 M PCl3 and 0.129 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the volume of the container is increased to 11.7 L? [PCl5] =  M [PCl3] =  M [Cl2] =  M

4. For the equilibrium CH₂(g) + 2H₂S(g) = CS₂(g) + 4H₂(g), the concentrations at equilibrium are [CH4] = 0.0804 M, [H₂S] = 0.0608 M, [CS₂] = 0.0196 M, and [H₂] = 0.0784 M at 1400.0 K. Using the equilibrium constant expression from before, calculate Kc.

Using the general properties of equilibrium constants 0/5 At a certain temperature, the equilibrium constant K for the following reaction is 7.4 × 10 ': N,(e) + 0,(g) -2NO(g) Use this information to complete the following table. Suppose a 35. L reaction vessel is filled with 1.7 mol of NO. What can you say about the composition of the mixture in the vessel at equilibrium? O There will be very little N, and 0,. O There will be very little NO. O Neither of the above is true. What is the equilibrium constant for the following reaction? Round your answer to 2 significant digits. K =] 2 NO(g) N2(0)+O2(a) 1, What is the equilibrium constant for the following reaction? Round your answer to 2 significant digits. K = ] 2N,(0)+20,(0) 4 NO(g) 1,