chem help please?

Question

Hydrogen sulfide decomposes according to the following reaction, for which Kc = 9.30 *10^-8 at 700°C.

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

If 0.48 mol H2S is placed in a 3.1 L container, what is the equilibrium concentration of H2(g) at 700°C? (answer in units of M)

Answer 1

Start by writing the expression for Kc:

Kc= [H2]^2[S2]/[H2S]^2

where ^2 means square. When you know Kc you always need molar concentrations, thus, for H2S we have:

[H2S]= 0.48 mol/3.1 L= 0.154839 L

Next, we set up a table as follows:

2 H2S(g) ----------------2 H2(g) ------------------- S2(g)
Initial: ----- 0.154839-------------- 0.0 ------------------------- 0.0 <--- Initial amounts
React: ------- 2X ------------------ 2X ------------------------ X --- <--- Amounts react
-------------------------------------------------------------------------------------------------------
At Equil: 0.154839-2X ---------- 2X ----------------------- X ---- <---- Amounts At Eq

Replacing the "At Equil" amounts into the expression for the equilibrium contant we have:

(2x)^2 (x) / (0.154839 -2x) =9.3 . 10^-8

Notice that this gives you a cubic equation (4x^3 in the numerator). While there are ways to solve a cubic equation we do not need to make out life miserable with this. We can simplify this equation and avoid the mess. For this let's look at the value of the equilibrium constant: 9.3 . 10^-8, this is a very small value. A small value of an equilibrium constant tells you that the reaction as written happens very little to the right. In other words, when you put H2S in a container at 700 C very little will be converted into H2 and S2. If this is so, we would expect the value of x to be very small compared to the initial concentration of H2S. Thus, we can make the following approximation:

0.154839 - 2x is the same as 0.154839 (because 2x is really, really small).

The above equation simplifies to:

(4x^3) / ( 0.15439) = 9.30 . 10^-8

This equation can be easily solved:

x = Cubic_Root(( 9.3 . 10^ -8. 0.15439) / 4)

which has the solution: X=8.23 . 10^-4

According to the table we wrote above the equilibrium concentrations are:

[H2S]=0.154839-2X= 0.153193

[H2] = 2X = 0.001646

[S2]= x = 0.000823

How do we check if our approximation is correct? The most common criterion is that if the value of the approximation (in this case 2x) is less than 5% of the value of the value on which we are approximating ( 0.154839) then it is ok. Let's see what percentage 2X is of 0.154839:

((2 x 0.000823.10^-4) / 0.154839) * 100 = 1.06 %

and we can see that our approximation was correct and that we can assume the above concentrations to be correct.

thus, out approximation is justified

Answer 2

Kc = ([H2]^2 * [S2])/[H2S]^2

The initial concentration of H2S is 0.48/3.1 M = 0.155 approx.

Suppose that 2x M of H2S decompose. We will have 2x M of H2 and x M of S2. To find x we have to solve the equation:

(2x)^2 * x / (0.155 - 2x)^2 = 9.3*10^-8

or

4x^3 / (0.155 - 2x)^2 = 9.3*10^-8

Mathematica gives x = 0.00153314 as a real root, so

[H2] = 2*0.00153314 = 0.003 M approximately