Which of the following reaction will get affected by increase of pressure? Also mention, whether change will cause the reaction to go into the right or left direction?
(`i`) `CH_(4(g))+2S_(2(g))hArrCS_(2(g))+2H_(2)S_((g))`
(`ii`) `CO_(2(g))+C_((s))hArr2CO_((g))`
(`iii`) `4NH_(3(g))+5O_(2(g))hArr4NO_((g))+6H_(2)O_((g))`
(`iv`) `C_(2)H_(4(g))+H_(2(g))hArrC_(2)H_(6(g))`

Answer

(i) unaffected;no shift (ii) affected; left direction.
(iii) affected; left (iv) affected; right

Explanation

(i) In this case `n_(p)` is equal to `n_(r)`
This reaction will not be affected by increase of pressure.
(ii) `CO_(2)(g)+C(s)hArr2CO(g)`
In this reaction, moles of gases on product side (`n_(p)=2`) is more than on reactant side (`n_(r)=1`). This reaction will be affected by increase in pressure. Increase in pressure shifts the equilibrium in that direction where there is less no. of moles gases. In this reaction increase in pressure will cause the reaction to go into the left direction.
(iii) `4NH_(3)(g)+5O_(2)(g)hArr4NO(g)+6H_(2)O(g)`
The reaction would be affected by increase in pressure because `N_(p)` is different form `n_(r)`
Increase in pressure shifts the equilibrium in left direction because `n_(t)` is less than `n_(p)`