Organic Chemistry/Introduction to reactions/Rates and equilibria

> Introduction to reactions

Chemical equilibria are ratios relating the forward and backward direction of a reaction to each other. This ratio is represented by the letter K in the following equation:

K = products / reactants

Definitions
Rate of reaction is the speed at which a chemical reaction takes place, expressed as moles per unit time and unit volume. The rate r of a general reaction $$aA + bB + ... -> p P + q Q + ...$$ is defined by:

$$ r = -{1 \over a} {dC_A \over dt} = -{1 \over b} {dC_B \over dt} = ... = {1 \over p} {dC_P \over dt} =.... $$

from the expression above, it is clear that the usual convention is that reaction rate is taken as products formation rate.

Rate is a function of the concentration of reactants and products, temperature, pressure and presence of a catalyst.

Rate expression
A common expression for reaction rate is the power law:

$$r = k C_A^ \alpha\ C_B ^ \beta\ ....$$

$$k$$ is called the kinetic constant, and $$ \alpha\ $$, $$ \beta\ $$, etc. are called the reaction order with respect to the reactant A, B (or the partial order of A, B etc.), respectively. The sum of all the orders is the global order of the rate expression. Therefore the rate expression $$ r = k C_A C_B $$ is a second-order and first-order in A and B.

The reaction orders are the same as the stoichiometric coefficient in the case of an elementary reaction only; in most cases they must be determined experimentally and are valid in the window of experimental conditions.

Elementary data on reaction orders can be obtained by changing the concentration C of one of the reactant, say A, and measuring the initial rate r. Plotting a rate vs concentration log-log graph, one obtains a straight line whose slope is the partial order in—say—A by virtue of the relation $$ log \, r = log \, k C_A ^ \alpha\ = log \, k + \alpha\ log \, C_A $$



Influence of temperature and pressure
$$ k = k_0 exp \left ( - \Delta\ G^{*} / RT \right)$$

as known as the Arrhenius equation

Rate equations and reaction mechanisms
The most important research application of kinetic investigations is the determination of reaction mechanisms. In fact, the rate expression is function of it.

From a postulated reaction mechanism (the model), a rate equation can be derived and used to analyse the experimental data. If the obtained fit is not statistically significant, the scheme is rejected. In complex systems, several schemes can produce compatible rate expressions and the problem of model discrimination is of primary importance.

Limiting step
Often a reaction has two or more steps. One of the steps, usually the last one, is the slowest step, and is said to be rate-limiting.

Steady state approximation
Sometimes it is useful, when calculating the reaction rate, to assume that no particular step is rate-limiting. Instead, the reaction intermediate can either proceed to the product or return to the original reactant with an equal rate for either possibility. This is called the steady-state approximation.

Example
A classical example is the hydrolysis of haloalkanes:


 * R-X + H2O &rarr; ROH + HX

This reaction can occur by two mechanisms: the SN1 and SN. The former is a unimolecular substitution: its rate is determined only by the concentration of R-X, without regard to the concentration of the new substituent. The latter is bimolecular: its rate is first-order in both R-X and new substituent, for a combined rate order of 2.

Equilibrium
Chemical equilibrium is the state when a net reaction is neither going forward nor backward. It is a dynamic equilibrium. The rate of the forward reaction equals the rate of the reverse reaction, so the two cancel each other out, and the net rate of change is zero.

The chemical equilibrium is dictated by the equilibrium constant (often written Keq), expressed by the mass-action law:

$$ K_{eq} = \frac { \prod_{i \in products} C_i}{ \prod_{j \in reactants} C_j } $$

The concentration C can be expressed in any scale, e.g. molar franction, molarity, partial pressure.

If the temperature and pressure are kept constant, no matter what are the initial concentrations, the system will evolve until the mass-action product is equal to the Keq. In general, systems with $$K_{eq} \ll 1$$ are highly displaced to the reactant sides (almost no conversion at equilibrium), whereas when $$K_{eq} \gg 1$$ the reaction goes to completion.

From classical thermodynamics it can be showed that the following relation holds:

$$K_{eq} = exp \left( -{\Delta\ G_R}/{RT} \right)$$

where $$\Delta\ G_R$$ is the total change in Gibbs free energy with reaction (product minus reactants).

Influence of temperature
The influence of temperature can be obtained by differentiation of the equation above to lead:

$$ -ln K_{eq} = \frac {\Delta\ H_R} {RT} - \frac {\Delta\ S_R}{R}$$

as known as the van't Hoff equation. Therefore, for exothermic reaction ($$\Delta\ H_r < 0 $$) an increase in temperature will decrease the $${\Delta\ H_R}/{RT}$$ quantity, leading to a lower Keq, conversely for an endothermic reaction.