\left. \begin{array} { l } { H _ { 2 } O ( g ) \rightarrow H _ { 2 } O

1.	\left. \begin{array} { l } { H _ { 2 } O ( g ) \rightarrow H _ { 2 } O ( 1 ) } \\ { H _ { 2 } ( g ) + l _ { 2 } ( g ) \rightarrow 2 H I ( g ) } \\ { C a C O _ { 3 } ( s ) \rightarrow C a O ( s ) + C O _ { 2 } ( g ) } \\ { N _ { 2 } ( g ) + 3 H _ { 2 } ( g ) \rightarrow 2 N H _ { 3 } ( g ) } \end{array} \right.
Answer» ∆s is positive is the no. of moles of gases formed in the product side will be more than that of the reactant. so, option 3 is the correct answer

\left. \begin{array} { l } { H _ { 2 } O ( g ) \rightarrow H _ { 2 } O ( 1 ) } \\ { H _ { 2 } ( g ) + l _ { 2 } ( g ) \rightarrow 2 H I ( g ) } \\ { C a C O _ { 3 } ( s ) \rightarrow C a O ( s ) + C O _ { 2 } ( g ) } \\ { N _ { 2 } ( g ) + 3 H _ { 2 } ( g ) \rightarrow 2 N H _ { 3 } ( g ) } \end{array} \right.

Answer»

∆s is positive is the no. of moles of gases formed in the product side will be more than that of the reactant.

so, option 3 is the correct answer