1.

Explain effect of catalyst on chemical equilibrium by example.

Answer»

Solution :In chemical reaction to increase or decrease the rate of reaction suitable catalyst is used. Catalyst does not affect the equilibrium. The value of K does not change by catalyst. Catalyst does not affect the COMPONENT of reaction mixture. The catalyst is not seen in equilibrium reaction or in expression of equilibrium constant.
Effect of catalyst: (i) A catalyst increases the rate of the chemical reaction by making available a new low energy pathway for the conversion of reactants to products. (ii) It increases the rate of forward and reverse reactions that pass through the same transition state and does not affect equilibrium, (iii) Catalyst lowers the activation energy for the forward and reverse reactions by exactly the same amount.
e.g.-1: Let us consider the formation of `NH_3` from `N_2` and `H_2` which is highly exothermic reaction and proceeds with decrease in total number of moles formed as COMPARED to the reactants.
`N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g))`
Equilibrium constant decreases with increase in temperature. At low temperature rate decreases and it takes long time to reach at equilibrium.
German chemist, Fritz Haber discovered that a catalyst consisting of iron catalyse the reaction to OCCUR at a satisfactory rate at temperatures `(500^@ C)`, where the equilibrium concentration of `NH_3` is reasonably favourable.
SINCE the number of moles formed in the reaction is less than those of reactants, the yield of `NH_3` can be improved by increasing the pressure.
Optimum conditions of temperature and pressure for the synthesis of `NH_3` using catalyst are around `500^@C` and 200 atm.
e.g.-2In manufacture of sulphuric acid by contact process,
`SO_(2(g)) + O_(2(g)) hArr 2SO_(3(g)) K_c=1.7xx10^26`
Though the value of `K_c` is suggestive of reaction going to completion, but practically the OXIDATION of `SO_2` to `SO_3` is very slow. Thus, platinum or divanadium penta-oxide `(V_2O_5)` is used as catalyst to increase the rate of the reaction.


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