consider the reaction. h2o(g)+cl2o(g)⇌2hclo(g) at equilibrium, the concentrations of the different species…

consider the reaction. h2o(g)+cl2o(g)⇌2hclo(g) at equilibrium, the concentrations of the different species are as follows. h2o=0.077 m cl2o=0.077 m hclo=0.023 m what is the equilibrium constant for the reaction at this temperature? 0.089 0.26 3.9 11

consider the reaction. h2o(g)+cl2o(g)⇌2hclo(g) at equilibrium, the concentrations of the different species are as follows. h2o=0.077 m cl2o=0.077 m hclo=0.023 m what is the equilibrium constant for the reaction at this temperature? 0.089 0.26 3.9 11

Answer

Answer:

0.089

Explanation:

Step1: Write equilibrium - constant expression

For the reaction $H_2O(g)+Cl_2O(g)\rightleftharpoons 2HClO(g)$, the equilibrium - constant expression $K_c=\frac{[HClO]^2}{[H_2O][Cl_2O]}$.

Step2: Substitute the given concentrations

Substitute $[H_2O]=0.077\ M$, $[Cl_2O]=0.077\ M$, and $[HClO]=0.023\ M$ into the expression: $K_c=\frac{(0.023)^2}{0.077\times0.077}$.

Step3: Calculate the value of $K_c$

$K_c=\frac{0.023^2}{0.077^2}=\left(\frac{0.023}{0.077}\right)^2\approx(0.299)^2\approx0.089$.