Respuesta :
Answer: The value of the equilibrium constant for this reaction is [tex]2.8\times 10^{-5}[/tex]
Explanation:
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
Moles of [tex]Ni(CO)_4[/tex] = [tex]\frac{0.597g}{170.7g/mol}=0.0035moles[/tex]
Moles of [tex]Ni[/tex] = [tex]\frac{12.7g}{58.7g/mol}=0.216moles[/tex]
Moles of [tex]CO[/tex] = [tex]\frac{1.98g}{28.01g/mol}=0.071moles[/tex]
Volume of solution = 2.7 L
Equilibrium concentration of [tex]Ni(CO)_4[/tex] = [tex]\frac{0.0035mol}{2.7L}=1.29\times 10^{-3}M[/tex]
Equilibrium concentration of [tex]Ni[/tex] = [tex]\frac{0.216mol}{2.7L}=0.08M[/tex]
Equilibrium concentration of [tex]CO[/tex] = [tex]\frac{0.071mol}{2.7L}=0.026M[/tex]
The given balanced equilibrium reaction is,
[tex]Ni(CO)_4\rightleftharpoons Ni+4CO[/tex]
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[Ni]^1\times [CO]^4}{[Ni(CO_4]^1}[/tex]
Now put all the given values in this expression, we get :
[tex]K_c=\frac{(0.08)^1\times (0.026)^4}{(1.29\times 10^{-3})^1}[/tex]
[tex]K_c=2.8\times 10^{-5}[/tex]
Thus the value of the equilibrium constant for this reaction is [tex]2.8\times 10^{-5}[/tex]
