The following reaction has an equilibrium constant of 0.50 at a given temperature.
HCHO(g) CO(g) + H2(g)
If you have 0.55 moles of HCHO(g) in a 1.0L container initially, how many moles of HCHO will be present when the system reaches equilibrium?
.5 = x*x/(.55-x)
solve for x, then .55-x
To determine the moles of HCHO present at equilibrium, we need to use the stoichiometry of the reaction and consider the equilibrium constant.
In the given reaction: HCHO(g) -> CO(g) + H2(g)
According to the stoichiometry, the reaction ratio is 1:1:1. This means that for every one mole of HCHO that reacts, one mole of CO and one mole of H2 are produced.
At equilibrium, let's assume that x moles of HCHO have reacted. Then, the equilibrium concentrations can be expressed as follows:
[HCHO] = (0.55 - x) moles/L
[CO] = x moles/L
[H2] = x moles/L
Using the equilibrium constant (Kc = [CO][H2]/[HCHO]), we can substitute the concentrations into the equation:
0.50 = (x)(x) / (0.55 - x)
This equation is called the quadratic equation and can be solved to find the value of x. However, solving this equation can be quite complicated.
Alternatively, we can make an approximation assuming that the initial amount of HCHO is significantly larger than the change at equilibrium. In this case, we can assume that (0.55 - x) ≈ 0.55 moles/L. By applying this approximation, we can simplify the equation to:
0.50 = (x)(x) / (0.55)
Now, we can solve for x:
0.50 * 0.55 = x^2
0.275 = x^2
Taking the square root of both sides, we get:
x = √0.275
x ≈ 0.524 moles/L
Therefore, when the system reaches equilibrium, approximately 0.524 moles of HCHO will be present.