Answer:
K = 4.00x10⁻³
Explanation:
Ni(s) + 4CO(g) → Ni(CO)₄(g)
Where equilibrium constant, K, is defined as:
K₁ = [Ni(CO)₄] / [CO] = 62599.3
The inverse reaction, has a K of:
Ni(CO)₄(g) → Ni(s) + 4CO(g)
K₂ = [CO] / [Ni(CO)₄]
K₂ = 1/K₁ = 1 / 62599.3 = 1.59746x10⁻⁵
Now, if reaction with the reaction decreases its coefficients in the half, K is:
¹/₂ Ni(CO)₄(g) → ¹/₂Ni(s) + 2 CO(g)
K₃ = √[CO] / √[Ni(CO)₄] = √K₂ = √1.59746x10⁻⁵ = 4.00x10⁻³
Answer:
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
