- #1

mishrashubham

- 599

- 1

I am aware that the equilibrium constant, K

[itex] {aA}~~+~~{bB}~~~\mathop{\rightleftharpoons}^{k_f}_{k_b}~~~{cC}~~+~~{dD}[/itex] is

[itex]\frac{k_f}{k_b} = \frac{[A]^a

_{c}for the reaction[itex] {aA}~~+~~{bB}~~~\mathop{\rightleftharpoons}^{k_f}_{k_b}~~~{cC}~~+~~{dD}[/itex] is

[itex]\frac{k_f}{k_b} = \frac{[A]^a

**^b}{[C]^c [D]^d}[/itex]**

Now my question is, are the stoichiometric coefficients a,b,c and d, written in simplest whole number ratios? Because if they can be written as we like, then the exponents of the concentrations of the reactants and products in the formula for the equilibrium constant will be different and we will have different values for the same reaction. My textbook says the relationship between the equilibrium constants for the two reactions

[itex] {aA}~~+~~{bB}~~~\mathop{\rightleftharpoons}^{k_f}_{k_b}~~~{cC}~~+~~{dD}[/itex] and

[itex] {naA}~~+~~{nbB}~~~\mathop{\rightleftharpoons}^{k'_f}_{k'_b}~~~{ncC}~~+~~{ndD}[/itex] is

[itex]\frac{k'_f}{k'_b} = \bigg(\frac{k_f}{k_b}\bigg)^n[/itex]

What I don't understand is whether it is trying to tell that the stoichiometric coefficients of a reaction should be converted into a simple whole number ratio before finding out the equilibrium constant or whether they will be different values altogether.

EDIT: I don't know why I can't get the rate constants above and below the harpoons but hope you get the point.Now my question is, are the stoichiometric coefficients a,b,c and d, written in simplest whole number ratios? Because if they can be written as we like, then the exponents of the concentrations of the reactants and products in the formula for the equilibrium constant will be different and we will have different values for the same reaction. My textbook says the relationship between the equilibrium constants for the two reactions

[itex] {aA}~~+~~{bB}~~~\mathop{\rightleftharpoons}^{k_f}_{k_b}~~~{cC}~~+~~{dD}[/itex] and

[itex] {naA}~~+~~{nbB}~~~\mathop{\rightleftharpoons}^{k'_f}_{k'_b}~~~{ncC}~~+~~{ndD}[/itex] is

[itex]\frac{k'_f}{k'_b} = \bigg(\frac{k_f}{k_b}\bigg)^n[/itex]

What I don't understand is whether it is trying to tell that the stoichiometric coefficients of a reaction should be converted into a simple whole number ratio before finding out the equilibrium constant or whether they will be different values altogether.

EDIT: I don't know why I can't get the rate constants above and below the harpoons but hope you get the point.

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