Medium
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of $ICl$ was $0.78 \, M$?
$2ICl_{(g)} \leftrightarrow I_{2(g)} + Cl_{2(g)}; \, K_c = 0.14$
$2ICl_{(g)} \leftrightarrow I_{2(g)} + Cl_{2(g)}; \, K_c = 0.14$
(N/A) The given reaction is:
$2ICl_{(g)} \leftrightarrow I_{2(g)} + Cl_{2(g)}$
Initial concentration: $[ICl] = 0.78 \, M$,$[I_2] = 0 \, M$,$[Cl_2] = 0 \, M$
At equilibrium: $[ICl] = (0.78 - 2x) \, M$,$[I_2] = x \, M$,$[Cl_2] = x \, M$
The equilibrium constant expression is:
$K_c = \frac{[I_2][Cl_2]}{[ICl]^2} = 0.14$
Substituting the values:
$\frac{x \cdot x}{(0.78 - 2x)^2} = 0.14$
Taking the square root on both sides:
$\frac{x}{0.78 - 2x} = \sqrt{0.14} \approx 0.374$
$x = 0.374(0.78 - 2x)$
$x = 0.2917 - 0.748x$
$1.748x = 0.2917$
$x \approx 0.167 \, M$
Therefore,at equilibrium:
$[I_2] = [Cl_2] = 0.167 \, M$
$[ICl] = 0.78 - 2(0.167) = 0.446 \, M$
$2ICl_{(g)} \leftrightarrow I_{2(g)} + Cl_{2(g)}$
Initial concentration: $[ICl] = 0.78 \, M$,$[I_2] = 0 \, M$,$[Cl_2] = 0 \, M$
At equilibrium: $[ICl] = (0.78 - 2x) \, M$,$[I_2] = x \, M$,$[Cl_2] = x \, M$
The equilibrium constant expression is:
$K_c = \frac{[I_2][Cl_2]}{[ICl]^2} = 0.14$
Substituting the values:
$\frac{x \cdot x}{(0.78 - 2x)^2} = 0.14$
Taking the square root on both sides:
$\frac{x}{0.78 - 2x} = \sqrt{0.14} \approx 0.374$
$x = 0.374(0.78 - 2x)$
$x = 0.2917 - 0.748x$
$1.748x = 0.2917$
$x \approx 0.167 \, M$
Therefore,at equilibrium:
$[I_2] = [Cl_2] = 0.167 \, M$
$[ICl] = 0.78 - 2(0.167) = 0.446 \, M$
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