At equilibrium a vessel at a constant total pressure of 1 atm contains 4 mol PC l 5 , 2 mol PC l 3 and 2 mol C l 2 . 15 mol of He is added at constant pressure and equilibrium is re-established.

Question

At equilibrium a vessel at a constant total pressure of 1 atm contains 4 mol PC l 5 ​ , 2 mol PC l 3 ​ and 2 mol C l 2 ​ . 15 mol of He is added at constant pressure and equilibrium is re-established. The number of moles of PC l 5 ​ at the new equilibrium is

TestHub · Accepted Answer

Initial K p ​ = ( 2/8 ) 2 / ( 4/8 ) × 1 = 1/8 atm . Let x mol more PC l 5 ​ dissociate. Total moles = 8 + 15 + x = 23 + x . K p ​ = ( 2 + x ) 2 / (( 23 + x ) ( 4 − x )) = 1/8 , so 8 ( 4 + 4 x + x 2 ) = 92 − 19 x − x 2 . 9 x 2 + 51 x − 60 = 0 , i.e. 3 x 2 + 17 x − 20 = 0 , so ( 3 x + 20 ) ( x − 1 ) = 0 and x = 1 . PC l 5 ​ = 4 − 1 = 3 mol . Trap: 4 mol (assuming He has no effect, as it would if Δ n = 0 ). Answer: (A).

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