# Chemistry

posted by .

Given the equation
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 -

I 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 -

So 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 -

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 -

Yeah 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 -

First, 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 -

Oh 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 :)

## Similar Questions

1. ### science

why is the unit Kelvin (temperature unit) not called "degree Kelvin"?
2. ### chemistry

Calculate the value of the equilibrium constant, Kp , for the following reaction at 298.0 Kelvin. (Use the reaction free energy given below.) 2SO3 = 2SO2 + O2 in the gaseous state ΔG = 140.0 kJ/mol
3. ### chemistry

Calculate the value of the equilibrium constant, Kp , for the following reaction at 298.0 Kelvin. (Use the reaction free energy given below.) 2SO3 = 2SO2 + O2 in the gaseous state ΔG = 140.0 kJ/mol
4. ### Chemistry

Given the equation NO(g) + 1/2)O^2(g) <-> NO(g) at 457 Kelvin with a K = 7.5 x 10^2 Find the K value for the equation 2NO^2(g) <-> 2NO(g) + O(g)

I have this question: Which is the smallest unit of measurement for temperature?
6. ### chemistry

calculate the rate of a constant reaction at 293 kelvin when the energy of activation is 103 kilo joules per mole and the rate constant at 273 kelvin is 7.87*10 to the power -7...
7. ### chemistry

calculate the rate of a constant reaction at 293 kelvin when the energy of activation is 103 kilo joules per mole and the rate constant at 273 kelvin is 7.87*10 to the power -7...
8. ### Math, check my answer

Kelvin and Marsha are going to dinner and a movie this evening. Kelvin wants to have at least \$70 cash in his wallet. He currently has \$10. How much money, x, does Kelvin need to withdraw from the bank?
9. ### physical science

The density of water is the greatest at a temperature of A. 0 Kelvin B. 277 Kelvin C. 273 Kelvin D. 4 Kelvin
10. ### Algebra

In the metric system there are two official temperature scales: degree Celsius and kelvin. The kelvin temperature scale is obtained by shifting the Celsius scale so of any heat whatsoever. An equation about the relationship between …

More Similar Questions