A gas X dissociates according to the reaction X ( g ) 2 Y ( g ) at a constant temperature. The degree of dissociation is very small compared to 1 ( 1 ). If the total equilibrium pressure is P , what is the slope of the linear plot obtained when ( ) is graphed against (P) ?

Step-by-step solution

For the reaction X ( g ) 2 Y ( g ) , let the initial moles of X be 1 . At equilibrium, moles of X = 1 - and moles of Y = 2 . Total moles at equilibrium = 1 - + 2 = 1 + . The partial pressures are: P_ X = ( 1- 1+ )P P_ Y = ( 2 1+ )P The equilibrium constant K_p is given by: K_p = (P_ Y )^2 P_ X = ( 2 1+ )^2 P^2 ( 1- 1+ )P = 4 ^2 P 1- ^2 Given that 1 , we can approximate 1 - ^2 1 . Thus, K_p 4 ^2 P . Rearranging for , we get: ^2 = K_p 4P = ( K_p 4 )^ 1/2 P^ -1/2 Taking the natural logarithm on both sides: ( ) = 1 2 (