Consider the following equilibrium process at 686 C

C02(g)+H2(g)=CO(g)+H20(g)
The equilibrium concentration of the reacting species are [CO]=0.050M, [H2]=0.045M, [CO2]=0.086M, and [H20]=0.040M. (a)Calculate the Kc for the reaction at 686 C. (b) If we add CO2 to increase its concentration to 0.50 mol/L, what will the concentration of all the gases be when equilibrium is reestablished.

I got (a) Kc= 0.52 but I'm having hard time getting part (b) because the answers are way different then what I'm getting. The answers should be:
[CO]=0.075M, [H2]=0.020M, [CO2]=0.48M, [H20]=0.065M.

At certain temperature the following reactions have the constant shown:

S(s) + O2(g) = SO2(g)
K'c=4.2*10^52

2S(s) + 3O2(g) = 2SO3(g)
K"c=9.8*10^128

Calculate the equilibrium constant Kc for the following reaction at that temperature:

2S02(s) + O2(g) = 2SO3(g)

This one was specially hard because time to time my calculator couldn't do it because of overflow.

Thanks.

I worked part b and obtained the answers you gave as correct.

(CO)(H2O)/(CO2)(H2) =
CO2 = 0.5-x
(H2O) = 0.045-x
(CO( = 0.05+x
(H2O) = 040+x
Solve for x, THEN add or subtract from the starting point to arrive at the equilibrium amounts. Post your work if you can't find the error and I'll give it a go.

I have just looked at the second problem. Is that SO2(s) a typo? Must be, huh?

ASSUMING that SO2(s) is a typo, then multiply equation 1 by 2 and reverse it. That new k1 = (1/k^2) and multiply that by k2.

That will be, if I haven't goofed in the arithmetic, is
k2/k1^2 = 9.8 x 10^128/(4.2 x 10^52)^2 = ??
You need not have a problem with overflow. Just do 9.8/(4.2)^2 on your calculator to obtain 0.555 or so and you do the exponent on paper. It is 128-52-52 = 24 (or 5.55 x 10^23). Check my work carefully.

Oh yes it is a typo my bad. But you got the right answer though. Thank you and thank you for the short cut on the calculator too, I was doing it just like you did it but my the calculator was having overflow problem

Thank you so much for your help!!!

(a) To calculate the equilibrium constant Kc for the given reaction at 686°C, we can use the given equilibrium concentrations of the reacting species. The general expression for the equilibrium constant Kc is:

Kc = ([CO] * [H2O]) / ([CO2] * [H2])

Given: [CO] = 0.050M, [H2] = 0.045M, [CO2] = 0.086M, [H2O] = 0.040M

Substituting these values into the equation:

Kc = (0.050 * 0.040) / (0.086 * 0.045)
= 0.025 / 0.00387
= 6.46

Therefore, the equilibrium constant Kc for the reaction at 686°C is approximately 6.46.

(b) To determine the concentrations of all the gases when equilibrium is reestablished after adding CO2 to increase its concentration to 0.50 mol/L, we need to consider the changes in concentrations of the reacting species. Let's assume x mol/L of CO2 is consumed during the reaction to establish equilibrium.

Using the stoichiometric coefficients, the new equilibrium concentrations can be expressed as:

[CO] = 0.050 - x
[H2] = 0.045 - x
[CO2] = 0.086 - x
[H2O] = 0.040 + x

Substituting these expressions into the equilibrium constant expression:

Kc = ([CO] * [H2O]) / ([CO2] * [H2])

6.46 = ((0.050 - x) * (0.040 + x)) / ((0.086 - x) * (0.045 - x))

This equation is quadratic in nature. By solving this equation, we can find the value of x (the change in concentration of CO2) and then calculate the final equilibrium concentrations of the gases.

Unfortunately, you have not provided the initial concentration of CO2, which is required to solve the equation and determine the value of x. Therefore, it is not possible to calculate the final equilibrium concentrations without that information.

Now, moving on to the second question:

Given reactions:
1. S(s) + O2(g) = SO2(g) (K'c = 4.2 * 10^52)
2. 2S(s) + 3O2(g) = 2SO3(g) (K''c = 9.8 * 10^128)

We need to calculate the equilibrium constant Kc for the following reaction:

2SO2(g) + O2(g) = 2SO3(g)

To obtain the equilibrium constant Kc for the desired reaction, we can use the relationship between the equilibrium constants of the given reactions:

Kc = (K''c) ^(2/3) / (K'c)

Substituting the given values:

Kc = (9.8 * 10^128) ^(2/3) / (4.2 * 10^52)

Evaluating this expression, we get:

Kc = (7.12 * 10^85) / (4.2 * 10^52)
= 1.70 * 10^33

Therefore, the equilibrium constant Kc for the reaction 2SO2(g) + O2(g) = 2SO3(g) at the given temperature is approximately 1.70 * 10^33.

can someone help me with part A of this question?

Thank you so much