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Will be there any change in the $K_c$ values for same reaction, if we write stoichiometry as a fraction or as a integer? For example, assuming that we calculate $K_c$ at same temperature:

$$\begin{align} &(1) &&\ce{1/2A + 3/2B -> C}\\ &(2) &&\ce{A + 3B -> 2C}\\ \end{align}$$

If we calculate $K_c$ for above two equilibrium reactions, do we get different answers?

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$$K_c = \mathrm{products \over reactants}$$ So this is done by multiplying the concentrations of the products and dividing by the concentrations of the reactants. The concentrations are put to the power of their coefficient. So in your problems at equilibrium $K_c$ would be: \begin{align} K_c &= \frac{[\ce{C}]^1}{[\ce{A}]^{\frac{1}{2}}[\ce{B}]^{\frac{3}{2}}}\tag1\\ K_c &= \frac{[\ce{C}]^2}{[\ce{A}]^1[\ce{B}]^3}\tag2\\ \end{align}

Note that $[\ce{A}]$ means concentration of $\ce{A}$.

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$K_c$ is a constant for a reaction at a given temperature, and these reactions ar equivalent, so $K_c$ is still constant. However if reaction (1) were at equilibrium and the concentration of all the reactants doubled (e.g. the pressure doubles or volume halves) the reaction will no longer be in equilibrium. The expression (2) with values multiplied by 2 will no longer equal $K_c$ and the equilibrium position will shift so that the value of the expression equals the equilibrium constant.

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    $\begingroup$ What perhaps needs to be added to this answer is that the law of mass action is defined for an equation int which the substances are in their lowest whol number ratios of stoichiometric coefficients. $\endgroup$ – Ben Norris Feb 27 '16 at 19:00

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