Equilibrium Constant Keq: 2H2S(g) <-> H2(g) + S2(g) at 830°C

[Fusilli_Jerry89]

The value of Keq for the reaction: 2H2S(g) <-> 2H2(g) + S2(g) is 4.20x10^-6 at 830 degrees celsius.
a) What is favoured in this reaction, reactants or products?
b) What concentration of S2 can be expected at equilibrium after 0.200 mol of H2S is injected into an empty 1.00 L flask?

 

[therealkellys]

a) Do you know how to write out Keq? (Hint: Use the law of mass action).
b) Make an ice box
2H2S(g) <--> 2H2(g) + S2 (g)
i .200 0 0
c -2x +2x +x
e .200 - 2x 2x x

Keq = [((2x)^2)(x)] / [(.200 - 2x)^2]

 

[Fusilli_Jerry89]

i solved x to equal 1.431 but when u plug it back into the original question, H2S comes out to a negative number

 

[Fusilli_Jerry89]

oh nm i made a mistake

 

[Fusilli_Jerry89]

k got down to 4x^3-1.68x10^-5x2+3.36x10^-6-1.68x10^-7=0 now what

 

[geoffjb]

k got down to 4x^3-1.68x10^-5x2+3.36x10^-6-1.68x10^-7=0 now what

What does that mean?

Look to your rules of equilibrium to determine which side of the reaction is favoured.

As for x, you can solve for it.

 

[Fusilli_Jerry89]

i put the products are favoured, but i do not know how to solve this equation because it has an 4^3 x^2 and an x

 

[geoffjb]

i put the products are favoured, but i do not know how to solve this equation because it has an 4^3 x^2 and an x

Why do you have $4x^{3}$ in your equation? Solve for x.