At a certain temperature, the equilibrium
constant Kc is 0.154 for the reaction
2 SO2(g) + O2(g) *) 2 SO3(g)
What concentration of SO3 would be in
equilibrium with 0.250moles of SO2 and 0.578
moles of O2 in a 1.00 liter container at this
temperature? Note: These latter moles are
the equilibrium values.
Answer in units of M.
2SO2 + O2 ==> 2SO3
K = 0.154 = (SO3)^2/(SO2)^2(O2)
Plug in SO2 and O2 equilibrium values and solve for the only unknown (SO3) you have.
I got .3335640868, it said I was wrong.
2SO2 + O2 ==> 2SO3
K = 0.154 = (SO3)^2/(SO2)^2(O2)
0.154 = (SO3)^2 / (0.250)^2 * 0.578
0.006843375 = (SO3)^2
(SO3) = 0.082724694
To find the concentration of SO3 at equilibrium, you need to set up an ICE table and apply the equilibrium expression.
Let's start by writing the balanced equation for the reaction:
2 SO2(g) + O2(g) → 2 SO3(g)
Now, let's set up the ICE table:
2 SO2(g) + O2(g) → 2 SO3(g)
----------------------------------------------
Initial: 0.250M 0.578M 0M
Change: -2x -x +2x
Equilibrium: 0.250-2x 0.578-x 2x
The equilibrium concentrations are given by the expression in the last row. We need to find the value of x, which represents the concentration of SO3 at equilibrium.
The equilibrium expression is given by:
Kc = (SO3)^2 / (SO2)^2 * (O2)
Plugging in the equilibrium concentrations:
0.154 = (2x)^2 / (0.250-2x)^2 * (0.578-x)
Now, solve the equation for x. Rearrange the equation and solve for x:
0.154 * (0.250-2x)^2 * (0.578-x) = (2x)^2
0.154 * (0.250-2x)^2 * (0.578-x) = 4x^2
Now, simplify and solve for x. This equation is a quadratic equation, so you will need to use the quadratic formula or a calculator to find the value of x. Once you find the value of x, substitute it back into the equilibrium expression to find the concentration of SO3 at equilibrium in Molarity (M).