Le chatelier's principle
If a system at equilibrium is subjected to a change of any one of the factors such as concentration, pressure or temperature, the system adjusts itself in such a way so as to minimise the effect of that change. Change of pressure: If a system in equilibrium consists of gases, then the concentrations of all the components can be altered by changing the pressure.When the pressure on the system is increased, the volume decreases proportionately.The total number of moles per unit volume will now be more and the equilibrium will shift in the direction in which there is decrease in number of moles i.e., towards the direction in which there is decrease in volume.
Effect of pressure on melting point : There are two types of solids :
(a)Solids whose volume decreases on melting e.g. ice, diamond , carborundum, magnesium nitride and quartz.
Solid (higher volume)`hArr` Liquid (lower volume)
The process of melting is facilitated at high pressure, thus melting point is lowered.
(b)Solids whose volume increase on melting, e.g. Fe,Cu, Ag,Au, etc.
Solid (lower volume)`hArr` Liquid (higher volume)
In this case the process of melting becomes difficult at high pressure, then melting point becomes high.
(c )Solubility of subtances : When solid substance are dissolved in water, either heat is evolved (exothermic) or heat is absorbed (endothermic).
`KCl+aq hArr KCl(aq)`-heat
In such cases,solubility increase with increase in temperature.
Consider the case of KOH, when this is dissolved heat is evolved.
`KOH+aq hArr KOH(aq)`+heat
In such cases, solubility decrease with increase in temperature.
(d)Solubility of gases in liquids : When a gas dissolves in liquid, there is decreases in volume. Thus, increases of pressure will favour the dissolution of gas in liquid.
Effect of temperature: Le-Chatelier's principle predicts a system at equilibrium will tend to shift in the endothermic direction when temperature is raised, for then energy is absorbed as heat and the rise in temperature is opposed.Conversely, an equilibrium will shift in the exothermic direction if the temperature is lowered, for then energy is released and the reduction in temperture is opposed Van't Hoff equation shows the dependence of equilibrium constant K on temperaure as :
`d/(dT)lnK=(DeltaH^@)/(RT^2) or ln K="constant"-(DeltaH^@)/R.1/T`
The plot of log K against `1/T` is a straight line with positive slope ( K being the equilibrium constant of a reaction ), which of the following is then correct ?

Options

A. The reaction is endothermic in nature
B. The reaction willl be exothermic in nature
C. The reaction goes to farther extent on raising the temperature
D. None of these

Answer

B

Explanation

Since the slope of the straight line graph between log K and `1/T` is positive ,`DeltaH_("reaction")` would be negative, i.e., the forward reaction would be exothermic.