Answer to Question #134613 in General Chemistry for Jennifer

Question

For the reaction 2HI D H2 + I2 (all gas phase) the value of the equilibrium constant at 700K is 0.0183. If 3.0 moles of HI are placed in a 5-liter vessel and allowed to decompose according to the above equation, what percentage of the original HI would remain at equilibrium?

Answer 1

Molarity of HI = 3 / 5 = 0.6 M . 

2HI = H₂+ I₂ .

0.6 M 0 0

0.6( 1 - x ) x x

0.0183 = x * x / 0.6 ( 1 - x ) = x ² / 0.6 .

0.0183 * 0.6 = x² .

x = 0.1047 M .

Remaining Concentration Of HI = 0.6 ( 1 - 0.1047 ) = 0.53718 M .

So , % Remaining Concentration of HI = 0.53718 * 100 / 0.6 = 89.53 % Answer .