Question

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

Answer 1

Answer: The value of the equilibrium constant for this reaction is $2.8\times 10^{-5}$

Explanation:

To calculate the moles, we use the equation:

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

Moles of  $Ni(CO)_4$ = $\frac{0.597g}{170.7g/mol}=0.0035moles$

Moles of  $Ni$ =  $\frac{12.7g}{58.7g/mol}=0.216moles$

Moles of  $CO$ =  $\frac{1.98g}{28.01g/mol}=0.071moles$

Volume of solution = 2.7 L

Equilibrium concentration of $Ni(CO)_4$ = $\frac{0.0035mol}{2.7L}=1.29\times 10^{-3}M$

Equilibrium concentration of $Ni$ = $\frac{0.216mol}{2.7L}=0.08M$

Equilibrium concentration of $CO$ = $\frac{0.071mol}{2.7L}=0.026M$

The given balanced equilibrium reaction is,

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

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[Ni]^1\times [CO]^4}{[Ni(CO_4]^1}$

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

$K_c=\frac{(0.08)^1\times (0.026)^4}{(1.29\times 10^{-3})^1}$

$K_c=2.8\times 10^{-5}$

Thus the value of the equilibrium constant for this reaction is $2.8\times 10^{-5}$