Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction the reaction to go into forward of backward direction.
a. `COCl_(2)(g) hArr CO(g)+Cl_(2)(g)`
b. `CH_(4)(g)+2S_(2)(g) hArr CS_(2)(g)+2H_(2)S(g)`
c. `CO_(2)(g)+C(s) hArr 2CO(g)`
d. `2H_(2)(g)+CO(g) hArr CH_(3)OH(g)`
e. `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g)`
f. `4NH_(3)(g)+5O_(2)(g) hArr 4NO(g)+6H_(2)O(g)`

Explanation

Reactions affected will be those in which `(n_(p) != n_(r))_(gaseous)` . Hence , reactions (i), (iii), (iv), (v) and (vi) will be affected . By applying Le Chetelier's principle , we can predict the direction . Increase of pressure will shift the equilibrium to the side in which the number of moles is less.
(i) `n_(p)=2, n_(r) = 1 , i.e, n_(p) gt n_(r) ,` reaction will go in the backward direction
(ii) `n_(p)=3, n_(r) = 3 , i.e, n_(p) gt n_(r) ,` reaction will not be affected by pressure .
(iii) `n_(p)=2, n_(r) = 1 , i.e, n_(p) gt n_(r) ,` reaction will go in the backward direction .
(iv) `n_(p)=1, n_(r) = 3 , i.e, n_(p) gt n_(r) ,` reaction will go in the fiorward direction .
(v) `n_(p)=1, n_(r) = 0 , i.e, n_(p) gt n_(r) ,` reaction wil go in the backward direction.
(vi) `n_(p)=10, n_(r) = 9 , i.e, n_(p) gt n_(r) ,` reaction will goo in the backward direction