Answer to Question #126333 in General Chemistry for Tyjaha

Question

For the reaction H2(g) + F2(g) ↔ 2 HF(g), the value of Kc = 195 at some temperature. The initial conditions are 0.50M HF and none of the LHS materials. Decide in which direction the reaction will spontaneously change. Then use the ICE scheme, set up properly, to solve for “x.” Once x is determined, fill in the bottom line to find all three equilibrium concentrations.

Answer 1

 H2(g) + F2(g) ↔ 2 HF(g), the value of Kc = 195 at some temperature

So 2 HF ↔ H₂ +F₂ , the value of Kc^, = (1/195 )at some temperature .

initial 0.50M 0 0

equillibrium 0.5(1- x) 0.5x 0.5x

 Kc^, = 0.5x * 0.5x / 0.5 ( 1-x) = 1 / 195 .

1- x = 1 ( approx ) .

( 0.5x )² = ( 0.5 ) * ( 1/295 ) = 0.001694

0.5x = [H₂] = [F₂] = 0.041158 .

FORMATION OF HF FROM H₂ AND F₂ ARE MORE favorable THEN dissociation OF HF TO H₂ AND F₂ .