Respuesta :
Answer:
The concentration of I at equilibrium = 3.3166×10⁻² M
Explanation:
For the equilibrium reaction,
I₂ (g) ⇄ 2I (g)
The expression for Kc for the reaction is:
[tex]K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}[/tex]
Given:
[tex]\left[I_2_{Equilibrium} \right][/tex] = 0.10 M
Kc = 0.011
Applying in the above formula to find the equilibrium concentration of I as:
[tex]0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}[/tex]
So,
[tex]\left[I_{Equilibrium} \right]^2=0.011\times 0.10[/tex]
[tex]\left[I_{Equilibrium} \right]^2=0.0011[/tex]
[tex]\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M[/tex]
Thus, The concentration of I at equilibrium = 3.3166×10⁻² M
A chemical reaction where the reactant splits into two more products is called a decomposition reaction. The decomposed iodine will have a concentration of [tex]3.3166 \times10^{-2} \;\rm M[/tex] at equilibrium.
What is the equilibrium constant for concentration?
The equilibrium constant for concentration [tex]\rm (K_{c})[/tex] is the proportion of the balanced concentrations of products with the reactants and their powers of stoichiometry.
The equilibrium reaction of decomposition can be shown as:
[tex]\rm I_{2} (g) \rightleftharpoons 2I (g)[/tex]
The equation for the [tex]\rm (K_{c})[/tex] is:
[tex]\rm K_{c} = \dfrac{[I]^{2}}{[I_{2}]}[/tex]
Given,
- Equilibrium concentration of [tex]\rm I_{2}[/tex] = 0.10 M
- Equilibrium constant [tex](\rm K_{c})[/tex] = 0.011
Substituting values in the above equation:
[tex]\begin{aligned}0.011 &= \dfrac{[\rm I]^{2}}{0.10}\\\\\rm [I_{\rm equilibrium}]^{2}&= 0.011 \times 0.10\\\\&= 3.31 \times 10^{-2} \;\rm M\end{aligned}[/tex]
Therefore, the concentration of I at equilibrium is [tex]3.3166 \times10^{-2} \;\rm M[/tex] .
Learn more about the equilibrium constant here:
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