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.