Answer :
Answer: The value of equilibrium constant for the given reaction is 99.85
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex] .....(1)
- For [tex]Fe^{3+}[/tex] ions:
Molarity of [tex]Fe^{3+}[/tex] solution = [tex]2.00\times 10^{-2}M[/tex]
Volume of solution = 7.00 mL = 0.007 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
[tex]2.00\times 10^{-2}M=\frac{\text{Moles of }Fe^{3+}\text{ ions}}{0.007L}\\\\\text{Moles of }Fe^{3+}\text{ ions}=(2.00\times 10^{-2}mol/L\times 0.007L)=1.4\times 10^{-4}mol[/tex]
- For [tex]SCN^{-}[/tex] ions:
Molarity of [tex]SCN^{-}[/tex] solution = [tex]1.50\times 10^{-3}M[/tex]
Volume of solution = 2.00 mL = 0.002 L
Putting values in equation 1, we get:
[tex]1.50\times 10^{-3}M=\frac{\text{Moles of }SCN^{-}\text{ ions}}{0.002L}\\\\\text{Moles of }SCN^-\text{ ions}=(1.50\times 10^{-3}mol/L\times 0.002L)=3\times 10^{-6}mol[/tex]
Volume of the container = [7 + 2 + 1] = 10 mL = 0.010 L
[tex]\text{Molarity of }Fe^{3+}\text{ ions}=\frac{1.4\times 10^{-4}mol}{0.01}=1.4\times 10^{-2}M[/tex]
[tex]\text{Molarity of }SCN^{-}\text{ ions}=\frac{3.0\times 10^{-6}mol}{0.01}=3.0\times 10^{-4}M[/tex]
The chemical equation for the formation of [tex][FeSCN^{2+}][/tex]complex follows:
[tex]Fe^{2+}+SCN^-\rightleftharpoons [FeSCN^{2+}][/tex]
Initial: 0.014 [tex]3.0\times 10^{-4}[/tex]
At eqllm: 0.014-x [tex]3.0\times 10^{-4}-x[/tex] x
We are given:
Equilibrium concentration of [tex][FeSCN^{2+}]=1.74\times 10^{-4}M=x[/tex]
Equilibrium concentration of [tex][Fe^{2+}]\text{ ions}=(1.4\times 10^{-2}-x)=(1.4-0.0174)\times 10^{-3}=1.383\times 10^{-2}M[/tex]
Equilibrium concentration of [tex][SCN^{-}]\text{ ions}=(3.0\times 10^{-4}-x)=(3.0-1.74)\times 10^{-4}=1.26\times 10^{-4}M[/tex]
The expression of [tex]K_{eq}[/tex] for above equation follows:
[tex]K_{eq}=\frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^-]}[/tex]
Putting values in above equation, we get:
[tex]K_{eq}=\frac{(1.74\times 10^{-4})}{(1.383\times 10^{-2})\times (1.26\times 10^{-4})}\\\\K_{eq}=99.85[/tex]
Hence, the value of equilibrium constant for the given reaction is 99.85