1) For the reaction system, 2 SO2(g) + O2(g) = 2 SO3(g), Kc has

a value of 4.62 at 450.0 K (Kelvin). A system, at equilibrium has the following concentrations:
(SO3) = 0.254 M; (O2) = 0.00855 M. What is the equilibrium concentration of SO2?

Is set this up as 4.62=(SO3)^2 / (SO2)^2 (O2)

4.62=(0.254)^2 / 0.00855(x^2)

4.62=0.064516/0.00855(x^2)

I am just having trouble doing the math now. I am not sure what to do.

x= sqrt (.064516/(.00855*4.62))

Put this in your google search window:

sqrt(.064516/(.00855*4.62))

0.500

To solve the equation, let's rearrange it to solve for (SO2)^2:

4.62 = (SO3)^2 / ((SO2)^2 * (O2))

We can simplify the equation further by multiplying both sides by ((SO2)^2 * (O2)):

4.62 * ((SO2)^2 * (O2)) = (SO3)^2

Now substitute the given values:

4.62 * ((SO2)^2 * 0.00855) = (0.254)^2

Simplify the equation:

0.0395013 * (SO2)^2 = 0.064516

Now divide both sides by 0.0395013 to isolate (SO2)^2:

(SO2)^2 = 0.064516 / 0.0395013

(SO2)^2 = 1.636486

Taking the square root of both sides gives us:

SO2 = √(1.636486)

Calculating the square root:

SO2 = 1.28 M

Therefore, the equilibrium concentration of SO2 is 1.28 M.

Thank you!!