Consider the following equilibrium at 1000K:
2SO2 (g) + O2 (g) ¡ê 2SO3 (g)
A study of this system reveals that there are 3.5E-3 moles of sulfur dioxide gas, and 4.8E-3 moles of oxygen gas present in a 11.0L flask at equilibrium. The equilibrium constant for this reaction is 4.2E-3. Calculate the number of moles of SO3(g) in the flask at equilibrium.
oops the equation is actually
2SO2 (g) + O2 (g) yield 2SO3 (g)
the yield part was an arrow facing both directions, not just the product side
To calculate the number of moles of SO3 gas (sulfur trioxide) in the flask at equilibrium, we can use the equilibrium constant (K) and the moles of the reactants present in the system.
The balanced equation for the reaction is:
2SO2 (g) + O2 (g) ↔ 2SO3 (g)
According to the information given, we have:
- Moles of SO2 (sulfur dioxide) = 3.5E-3 moles
- Moles of O2 (oxygen) = 4.8E-3 moles
- Equilibrium constant (K) = 4.2E-3
We can set up an ICE table to keep track of the changes in moles during the reaction:
```
2SO2 + O2 ↔ 2SO3
Initial: 3.5E-3 4.8E-3 0
Change: -2x -x +2x (since the stoichiometric ratio is 2:1:2)
Equilibrium: 3.5E-3 - 2x 4.8E-3 - x 2x
```
The equilibrium expression for this reaction is given by:
K = [SO3]^2 / ([SO2]^2 * [O2])
Since we know K = 4.2E-3, we can plug in the equilibrium concentrations into the expression and solve for x.
4.2E-3 = (2x)^2 / ((3.5E-3 - 2x)^2 * (4.8E-3 - x))
Simplifying and rearranging the equation:
4.2E-3 * (3.5E-3 - 2x)^2 * (4.8E-3 - x) = (2x)^2
This equation is quadratic, so we can solve it either graphically or using numerical methods such as iterative approximation or solving it using a computer program. The latter method is more reliable and efficient.
Therefore, we can use a numerical solver or software to find the value of x that satisfies the equation. The resulting value of x will give us the moles of SO3 in the flask at equilibrium.