Adam

When a 3.00 L flask is used, and 0.46 mol of H2 and 0.30 mol of S2 are present in the flask initially, what equilibrium concentrations will exist for all species at some later time?

2H2S(g) <--> 2H2(g) + S2(g)

I am not sure how to do this question, if someone can just guide me in the correct direction.

.... 2H2(g) + S2(g) <-->2H2S(g)

I .... 0.15.....0.1............0

C......-2x......-x.............+2x

E...0.15-2x...0.1-x..........2x

Using this set up I keep getting a higher concentration for x than what I started with, which is incorrect. What am I doing incorrectly?

Thanks

below occurs and the temperature of the resulting solution rises from 22.00C to 33.10C. If the specific heat of the solution is 418 J/(g*C) calculate delta H for the reaction, as written.

Ba(s)+2H2O(l)->Ba(OH)2(aq)+H2(g)

Now I know the answer is 431 kJ, I am just trying to understand why it is exothermic (-431), considering we gained heat. Can someone enlighten me please?

The gases then reach equilibrium according to the equation: 2SO2(g) + O2(g) <--> 2SO3(g)

At equilibrium, the bulb was found to contain 2.6 moles of sulfur dioxide.Calculate Keq for the reaction.

I got an answer of 130.47 as the constant. I used an ICE table and determined x=0.14 mol/L

....2SO2(g) + O2(g) <--> 2SO3(g)

I....0.4...........0.22...........0.56

C...-x...........-x..................2x

E..0.26.........0.08..............0.84

Just making sure this is correct.