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Before answering my question, take a look at this example

$$\ce{2NOCl <=> 2NO + Cl2}$$

Let's suppose I have 2 mol/l $\ce{NOCl}$, 2 mol/l $\ce{NO}$, and 1 mol/l $\ce{Cl2}$ in a closed system. If we were to calculate the equilibrium constant, it would be

$$K_c = \frac{(2)^2(1)}{(2)^2}$$

Now suppose I increase the concentration of both $\ce{NOCl}$ and $\ce{NO}$ by 1 mol/l. The equilibrium will stay the same,

$$K_c = \frac{(2+1)^2(1)}{(2+1)^2};$$

however, if I were to instead increase the concentration of both $\ce{NOCl}$ and $\ce{Cl2}$ by 1 mol/l, the equilibrium will shift to the product side

$$K_c = \frac{(2)^2(1+1)}{(2+1)^2}$$

Can someone explain this phenomenon in terms of rate of reaction or in terms of molecules?

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In an equilibrium, the equilibrium constant remains the same. When you change the concentration of some component, the whole system shifts to new concentrations to maintain this constant.

When one mole each of $\ce{NOCl}$ and $\ce{NO}$ are added, let us suppose that the equilibrium is shifted to the right and some $\ce{Cl2}$ is formed. The new concentration of $\ce{NOCl}$ will now be $3-x$, the concentration of $\ce{NO}$ will become $3+x$ and the one of $\ce{Cl2}$, $1+\frac{x}{2}$.

The final concentrations can be calculated by solving the cubic equation:

$$K_c=\frac{(3+x)^2 (1+\frac{x}{2})}{(3-x)^2}$$

Note that if in reality the equilibrium does shift to the left instead of the right, the value of $x$ will be negative.

In the other case, the equilibrium will shift to the left, and the equation to be solved is:

$$K_c=\frac{(2-x)^2 (1-\frac{x}{2})}{(3+x)^2}$$

Again, a negative value of $x$ will mean that the equilibrium shifts to the right instead of going to the left as has been supposed.

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Can someone explain this phenomenon in terms of rate of reaction or in terms of molecules?

In terms of rate of reaction, at equilibrium the forward rate is proportional to the concentrations of the reactant (as if this were an elementary reaction). Likewise, at equilibrium the reverse rate is proportional to the concentrations of the products (as this were an elementary reaction). If you change the concentration of a single reactant and a single product by the same factor (i.e. multiply by the same number, not add the same number) and if the stoichiometric factors are the same (i.e. the order of reaction in this special case), and the reactants are sufficiently dilute, the system will still be at equilibrium. In the second change described by the OP, neither is the case (one concentration increases by 50%, the other doubles, and the exponents are different).

In terms of thermodynamics, if the Gibbs free energy of the reactants is changed by the same amount (i.e. add or subtract a number) the system will remain at equilibrium. This will happen if one species on either side is changed by the same factor because the Gibbs free energy changes with the logarithm of the activity of a species. Again, this only works if the stoichiometric coefficient of the two species are the same.

Nomenclature K vs Q

In the equations shown by the OP, different concentrations are substituted into the equilibrium constant expression even when the reaction is not at equilibrium. The correct name for these expresssions is reaction quotient Q. If Q matches K, the system is at equilibrium. In the examples provided, Q stays the same in one case (still in equilibrium) and changes in the other case (no longer in equilibrium). What the equilibrium would be needs not to be calculated. Knowing that Q is unequal to K is sufficient for showing that the system is in fact not in equilibrium.

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When a reaction shifts to the right or left direction, it means that the reaction is trying to remove the equilibrium stress. So removing or adding reagent or product in a reaction at equilibrium will not change the value of the constant.

You said you would increase the amount of substance of $\ce{NOCl}$ and $\ce{NO}$, but you did not add it into $\ce{NO}$ concentration.

Adding products and reagents concentration would change the constant's value, but not shift the equilibrium towards any side.

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  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". Unfortunately your answer is a bit ambiguous, if not wrong. If the equilibrium doesn't shift, the constant's value stays the same, too. $\endgroup$ – Martin - マーチン Aug 27 '18 at 12:29
  • $\begingroup$ (-1) The equilibrium constant $K_c$ does not change... it is the reaction quotient $Q_c$ that changes. $\endgroup$ – orthocresol Aug 27 '18 at 12:39

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