# For a certain reaction 2A + B rightleftharpoons C + 3D, K_(eq) = 4.2 xx 10^3, which of the following is true?

## a) The reaction is product-favored. b) The reaction rate is fast. c) There are more reactants than products. d) The reaction is neither product-favored nor reactant-favored.

May 13, 2016

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#### Explanation:

The reaction lies to the right side because of the following:

The expression of the equilibrium constant $K$ is:

$K = \left(\left[\text{Products"])/(["Reactants}\right]\right)$

Since $K > 1$ this means that the $\left[\text{Products}\right]$ is greater than the $\left[\text{Reactants}\right]$ and thus there is more products than reactants, and therefore, the equilibrium lies to the right.

Chemical Equilibrium | Reaction Quotient & Application of a Large K.

May 13, 2016

The equilibrium constant for this reaction is:

"K"_(eq) = \frac(["products"])(["reactants"]) = \frac([C][D]^3)([A]^2[B]) = 4.2xx10^3

${\text{K}}_{e q} > 1$, thus $\left[\text{products"] > ["reactants}\right]$, and the equilibrium lies to the "right" (a), or the "products" side.

(b) We cannot say what the rate of reaction is, because that's a kinetic description of a reaction whose kinetic quantities are unstated. We only know the equilibrium constant, which is a thermodynamic quantity. $\Delta {G}^{\circ} = - R T \ln {K}_{e q}$.

(c) This is backwards. If you had more reactants than products, then $0 < {\text{K}}_{e q} < 1$, since $\left[\text{reactants"] < ["products}\right]$.

(d) This is only true if ${\text{K}}_{e q} = 1$, i.e. $\left[\text{reactants"] = ["products}\right]$.