Question

Nickel carbonyl decomposes to form nickel and carbon monoxide, like this:

Ni(CO)4 → Ni + 4CO

At a certain temperature, a chemist finds that a 2.7L reaction vessel containing a mixture of nickel carbonyl, nickel, and carbon monoxide at equilibrium has the following composition:

Compound Amount
Ni(CO)4 0.597g
Ni 12.7g
CO 1.98g

Calculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.

Answer 1

Answer: The value of the equilibrium constant for this reaction is
`2.8* 10^(-5)`

Step-by-step explanation:

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}`

Moles of
`Ni(CO)_4` =
`(0.597g)/(170.7g/mol)=0.0035moles`

Moles of
`Ni` =
`(12.7g)/(58.7g/mol)=0.216moles`

Moles of
`CO` =
`(1.98g)/(28.01g/mol)=0.071moles`

Volume of solution = 2.7 L

Equilibrium concentration of
`Ni(CO)_4` =
`(0.0035mol)/(2.7L)=1.29* 10^(-3)M`

Equilibrium concentration of
`Ni` =
`(0.216mol)/(2.7L)=0.08M`

Equilibrium concentration of
`CO` =
`(0.071mol)/(2.7L)=0.026M`

The given balanced equilibrium reaction is,

`Ni(CO)_4\rightleftharpoons Ni+4CO`

The expression for equilibrium constant for this reaction will be,

`K_c=([Ni]^1* [CO]^4)/([Ni(CO_4]^1)`

Now put all the given values in this expression, we get :

`K_c=((0.08)^1* (0.026)^4)/((1.29* 10^(-3))^1)`

`K_c=2.8* 10^(-5)`

Thus the value of the equilibrium constant for this reaction is
`2.8* 10^(-5)`