The equilibrium constant for the reaction Ni (s) + 4 CO (g)imageNi(CO)4 (g) is 62,599.3 at a certain temperature. What is the value of the equilibrium constant for the reaction below, at that same temperature? ½ Ni(CO)4 (g)image½ Ni(s) + 2 CO (g)

The equilibrium constant for the reaction  ½ Ni(CO)4 (g) ⇆ ½ Ni(s) + 2 CO (g) is 0.00400

Explanation:

Step 1: Data given

Ni (s) + 4 CO (g) ⇆ Ni(CO)4 (g)    Kc = 62599.3

Step 2: The balanced equation

½ Ni(CO)4 (g) ⇆ ½ Ni(s) + 2 CO (g)

Step 3: Calculate Kc for the reverse reaction

Ni(CO)4 (g) ⇆ Ni (s) + 4 CO (g)

Kc' = 1/Kc

Kc' = 1/ 62599.3

Kc' = 1.60*10^-5

Step 4: Calculate Kc for half of the reaction

½ Ni(CO)4 (g) ⇆ ½ Ni(s) + 2 CO (g)

Kc" = √(Kc')

Kc" = √(1.60*10^-5)

Kc" = 0.00400

The equilibrium constant for the reaction  ½ Ni(CO)4 (g) ⇆ ½ Ni(s) + 2 CO (g) is 0.00400

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.