Aluminum metal and bromine liquid (red) react violently to make aluminum bromide (white powder). One way to represent this equilibrium is: 2 Al(s) + 3 Br2(1)=2 AIBR3(s) We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above. 1) 2 AIBR3(s) =2 Al(s) + 3 Br2(1) K1 = 2) Al(s) + 3/2 Br2(1)=AIBR3(s) 3) AIBR3(s) K2 = 2 Al(s) + 3/2 Br2(1) K3 = Drag and drop your selection from the following list to complete the answer: 1/K (1/K)12 K12

In the context of equilibrium constants of chemical reactions, which "K" value indicates a reaction that favors the formation of products the most? K- 1.7 x 10-6 K= 8.2 x 10-3 OK- 5.31 x 10 OK- 4.99 x 106

write your answer in scientific notation

At a particular temperature, the equilibrium amounts of hydrogen, iodine and hydrogen iodide present in a 4.00 L flask are found to be: 0.25 moles of hydrogen gas, 0.15 moles of iodine gas and 1.37 moles of hydrogen iodide gas. H2(g) + I2(g)  ↔  2 HI(g) Calculate the equilibrium constant, Kc, value for the reaction.

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Suppose a 500. mL flask is filled with 0.50 mol of Cl₂, 1.7 mol of HCl and 0.20 mol of CC14. The following reaction becomes possible: Cl,(g)+CHCI, (g) → HCI(g) +CCI, (g) 4 The equilibrium constant K for this reaction is 0.104 at the temperature of the flask. Calculate the equilibrium molarity of Cl₂. Round your answer to two decimal places.

The reaction listed below is currently at equilibrium. If you were to suddenly halve the volume of the reaction chamber, what effect would that have on the system? Kr (g) + 2 F2 (3) = KrF4 (s) The reaction will shift to the left in the direction of reactants. O Cannot be determined - The equilibrium concentrations of all species must be provided. The reaction will shift to the right in the direction of products. O The equilibrium constant will increase. O The equilibrium constant will decrease.

Write the equilibrium constant expression, K, for the following reaction: (If either the numerator or denominator is blank, please enter 1.) CO(g) + Cl2 (g) 2 COCI2 (g) K=