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

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explanation:

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number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.