Tuesday

September 16, 2014

September 16, 2014

Posted by **K** on Saturday, May 28, 2011 at 2:32pm.

2H^2S(g) <-> 2H^2(g) + S^2(g)

With a Kp = 1.2x10^-2 at 1338 Kelvin

Find the value of Kc for the reaction

H^2(g) + 1/2S^2(g) <-> H^2S(g) at 1338 Kelvin

- Chemistry -
**DrBob222**, Saturday, May 28, 2011 at 5:08pmI would first convert Kp to Kc using Kp = Kc(RT)^delta n, then see your next post to adjust for the equation. It appears to me to be reversed and 1/2; therefore, K'c = 1/(sqrt Kc).

- Chemistry -
**K**, Saturday, May 28, 2011 at 5:35pmSo instead of looking at each compound in the equation you would only look at the product side of it and compare it to the other equation and only either square,square root etc. that value?

- Chemistry -
**DrBob222**, Saturday, May 28, 2011 at 5:53pmI'm not sure I follow your thinking.

Kp = Kc(RT)^dn

Kc = Kp(RT)^dn

Kc = 1.2E-2(0.08205*1338)^(3-2)

Kc = 1.2E-2(o.08205*1338) = ??

Then K'c = (1/sqrt Kc)

- Chemistry -
**K**, Saturday, May 28, 2011 at 6:02pmYeah I did all the previous things to find Kc and I got kc = 0.012. But you know how for to get the other kc you said I would have to do 1/sqrt Kc, I know that I would do that because the product is H2S and I would do the inverse because it is reversed and I would sqrt because it is halved. I get all of that but does that mean that we don't have to do anything to the reactants side of 1/2S2 and H2? Wouldn't we have to do the inverse of each of those since they are reversed as well and do the sqrt of those too as well since both are halved from the original equation?

- Chemistry -
**DrBob222**, Saturday, May 28, 2011 at 6:51pmFirst, let me correct an error I made in typing. I omitted the divisor sign.

Kp = Kc(RT)^dn

Kc = Kp(RT)^dn**This should be**

Kc = Kp/(RT)^dn

Kc = 1.2E-2(0.08205*1338)^(3-2)**This should be**

Kc = 1.2E-2/(0.08205*1338)^3-2

Kc = 1.2E-2(o.08205*1338) = ??**This should be**

Kc = 1.2E-2/(0.08205*1338) = ?? for which I obtained 1.1E-4.

Then K'c = (1/sqrt Kc)**Then 1/(sqrt 1.093E-4) =?? for which I obtained 95.6 which rounds to 96 to two s.f.**

Your confusion arises over K and your concept of what this procedure does.

K for the original equation is

Kc = (H2)^2(S2)/(H2S)^2 = 1.093E-4

When you reverse it, the equation becomes 2H2 + S2 ==> 2H2S and

Kc now is (H2S)^2/(H2^2)(S2) = 1/1.093E-4 = 9149.69 (You can see that the new K is the reciprocal of the old K from the way two K expressions are written.) And I am operating on K; for example, I don't take the reciprocal once for H2S, once for S and once for H2 and that is because the new K is just the reciprocal of the old K. Now when we go to 1/2, note we are taking 1/2 of H2S, 1/2 of S2 and 1/2 of 2H2 so the new K now is (H2S)/(H2)(S)^1/2. By taking the sqrt of K (the entire expression), we actually take the sqrt of (H2S)^2[it was (H2S)^2 and becomes (H2S)]; we take the sqrt of (H2)^2 [it was (H2)^2 and becomes (H2)]; we take the sqrt (S) [it was (S) and becomes sqrt S]/ I hope this helps. It's important that you understand what we've done. ;-).

- Chemistry -
**K**, Saturday, May 28, 2011 at 7:49pmOh okay, thank you so much for your whole explanation. I finally understand the whole concept behind it and what we are suppose to do. Thanks so much again :)

**Answer this Question**

**Related Questions**

Chemistry - Given the equation NO(g) + 1/2)O^2(g) <-> NO(g) at 457 Kelvin ...

Chemistry - In an experiment, 0.0300 moles each of SO3(g), S02(g), and O2(g) ...

chemistry - Calculate the value of the equilibrium constant, Kp , for the ...

chemistry - Calculate the value of the equilibrium constant, Kp , for the ...

Chem - I was given a problem that my book did not explain how to do, so I ...

Chemistry - A solution of aniline(C6H5NH2, Kb=4.2x10^-10) has a pH of 8.69 at ...

Chemistry - What is the pH of a 1.24 mol/L solution of HCN(aq) if its Ka = 6.2 x...

Chemistry - For the reaction given, the value of the equilibrium of the constant...

chemistry - given the equilibrium constant for the following reaction at 500 K, ...

chemistry - why should we add 273 to convert a given celcius value in to kelvin?