12) The K_(eq) for the equilibrium below is 7.52times 10-2 at 480.0^circ C 2Cl_(2)(g)+2H_(2)O(g)leftharpoons 4HCl(g)+O_(2)(g) What is the value of K_(eq) at this temperature for the following reaction? 2HCl(g)+(1)/(2)O_(2)(g)leftharpoons Cl_(2)(g)+H_(2)O(g) A) 133 B) 3.65 C) -0.0376 D) 5.66times 10-3 E) 0.274 Answer: B

To find the value of $$K_{eq}$$ for the given reaction, we need to manipulate the given equilibrium expression to match the desired reaction.

The given reaction is:
$$2Cl_{2}(g)+2H_{2}O(g)\rightleftharpoons 4HCl(g)+O_{2}(g)$$

The equilibrium constant expression for this reaction is:
$$K_{eq} = \frac{[HCl]^4[O_{2}]}{[Cl_{2}]^2[H_{2}O]^2}$$

Now, let's manipulate this expression to match the desired reaction:
$$2HCl(g)+\frac {1}{2}O_{2}(g)\rightleftharpoons Cl_{2}(g)+H_{2}O(g)$$

To do this, we need to reverse the reaction and change the stoichiometry. Reversing the reaction gives us:
$$4HCl(g)+O_{2}(g)\rightleftharpoons 2Cl_{2}(g)+2H_{2}O(g)$$

Now, we need to change the stoichiometry to match the desired reaction. We can do this by dividing the entire reaction by 2:
$$2HCl(g)+\frac {1}{2}O_{2}(g)\rightleftharpoons Cl_{2}(g)+H_{2}O(g)$$

The equilibrium constant expression for this manipulated reaction is:
$$K_{eq} = \frac{[Cl_{2}][H_{2}O]}{[HCl]^2[O_{2}]^{\frac{1}{2}}}$$

Now, we can substitute the given value of $$K_{eq}$$ for the original reaction into this expression:
$$K_{eq} = \frac{[Cl_{2}][H_{2}O]}{[HCl]^2[O_{2}]^{\frac{1}{2}}} = 7.52\times 10^{-2}$$

To find the value of $$K_{eq}$$ for the desired reaction, we need to take the square root of the given $$K_{eq}$$ value:
$$K_{eq} = \sqrt{7.52\times 10^{-2}} = 3.65$$

Therefore, the value of $$K_{eq}$$ for the desired reaction at $$480.0^{\circ }C$$ is 3.65, which corresponds to option B.