Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :
[butane]=1.0 M , [isobutane]=2.5 M
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?
Answer:
The equilibrium concentration of each gas:
[Butane] = 1.14 M
[isobutane] = 2.86 M
Explanation:
Butane ⇄ Isobutane
At equilibrium
1.0 M 2.5 M
After addition of 0.50 M of butane:
(1.0 + 0.50) M -
After equilibrium reestablishes:
(1.50-x)M (2.5+x)
The equilibrium expression will wriiten as:
x = 0.36 M
The equilibrium concentration of each gas:
[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M
[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M
To be able to write correctly the equilibrium expression of a reaction, we need to know the balanced reaction and the phases of the substances in the reaction. When substances are solid, pure liquid they are not included in the expression. We do as follows:
<span>4KO2(s) + 2H2O(g) = 4KOH(s) + 3O2(g)
K = [O2]^3 / [H2O]^2</span>
Sliding or rolling friction
Answer:
The mass number (A), also called atomic mass number or nucleon number, is the total number of protons and neutrons (together known as nucleons) in an atomic nucleus. The mass number is different for each different isotope of a chemical element.
This problem is providing the mass-volume percent of a formic acid solution so its molarity is required and found to be 0.313 M after the following calculations.
<h3>Molarity</h3>
In chemistry, units of concentration provide a measurable understanding of the relationship between the relative amounts of both solute and solvent. In the case of molarity, one must relate moles of solute and liters of solution as follows:
In such a way, when given this mass-volume percent of 1.44% for the formic acid in the solution, one can assume there is 100 mL of solution and 1.44 g of solute (formic acid), which means one must convert the volume to liters and the mass to moles with:
Finally, we plug in these numbers in the equation for the calculation of molarity:
Learn more about molarity: brainly.com/question/10053901