The question is incomplete, here is the complete question:
Nickel and carbon monoxide react to form nickel carbonyl, like this:
At a certain temperature, a chemist finds that a 2.6 L reaction vessel containing a mixture of nickel, carbon monoxide, and nickel carbonyl at equilibrium has the following composition:
Compound Amount
Ni 12.7 g
CO 1.98 g
0.597 g
Calculate the value of the equilibrium constant.
<u>Answer:</u> The value of equilibrium constant for the reaction is 2448.1
<u>Explanation:</u>
We are given:
Mass of nickel = 12.7 g
Mass of CO = 1.98 g
Mass of
= 0.597 g
Volume of container = 2.6 L
To calculate the number of moles for given molarity, we use the equation:
![\text{Molarity of the solution}=\frac{\text{Given mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20the%20solution%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%20of%20solute%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20solute%7D%5Ctimes%20%5Ctext%7BVolume%20of%20solution%20%28in%20L%29%7D%7D)
![\text{Equilibrium concentration of nickel}=\frac{12.7}{58.7\times 2.6}=0.083M](https://tex.z-dn.net/?f=%5Ctext%7BEquilibrium%20concentration%20of%20nickel%7D%3D%5Cfrac%7B12.7%7D%7B58.7%5Ctimes%202.6%7D%3D0.083M)
![\text{Equilibrium concentration of CO}=\frac{1.98}{28\times 2.6}=0.0272M](https://tex.z-dn.net/?f=%5Ctext%7BEquilibrium%20concentration%20of%20CO%7D%3D%5Cfrac%7B1.98%7D%7B28%5Ctimes%202.6%7D%3D0.0272M)
![\text{Equilibrium concentration of }Ni(CO)_4=\frac{0.597}{170.73\times 2.6}=0.00134M](https://tex.z-dn.net/?f=%5Ctext%7BEquilibrium%20concentration%20of%20%7DNi%28CO%29_4%3D%5Cfrac%7B0.597%7D%7B170.73%5Ctimes%202.6%7D%3D0.00134M)
For the given chemical reaction:
![Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)](https://tex.z-dn.net/?f=Ni%28s%29%2B4CO%28g%29%5Crightarrow%20Ni%28CO%29_4%28g%29)
The expression of equilibrium constant for the reaction:
![K_{eq}=\frac{[Ni(CO)_4]}{[CO]^4}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BNi%28CO%29_4%5D%7D%7B%5BCO%5D%5E4%7D)
Concentrations of pure solids and pure liquids are taken as 1 in equilibrium constant expression.
Putting values in above expression, we get:
![K_{eq}=\frac{0.00134}{(0.0272)^4}\\\\K_{eq}=2448.1](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B0.00134%7D%7B%280.0272%29%5E4%7D%5C%5C%5C%5CK_%7Beq%7D%3D2448.1)
Hence, the value of equilibrium constant for the reaction is 2448.1