The equilibrium constant expression for this reaction is
. This is an example of a heterogeneous equilibrium wherein the states of the reactants and products are different. In such a case, only the concentration of the gaseous and aqueous substances are included in the equilibrium constant expression.
Further Explanation:
The equilibrium constant expression is the ratio of the concentration of the products and the concentration of reactants.
The guidelines in writing the equilibrium constant expressions are as follows:
- Write molar concentration of each product in the numerator of the Keq expression.
- Write the coefficient of the substance as the exponent of the molar concentration of the substance in the Keq expression.
- Write the molar concentration of each reactant raised to its coefficient in the denominator.
Note: Pure substances (i.e. solids and liquids) are not included in the equilibrium constant expression as their concentrations are constant.
The numerical equivalent of the Keq expression is called the equilibrium constant. It has a specific value for a given temperature. The equilibrium constant provides information about the spontaneity and progress of an equilibrium reaction.
Learn More
- Learn more about equilibrium constants brainly.com/question/4137132
- Learn more about Le Chatelier's Principle brainly.com/question/12983923
- Learn more about Gibbs Free Energy brainly.com/question/12979420
Keywords: equilibrium constant expression, equilibrium
Answer:
Empirical formula is C₂H₆O.
Explanation:
Empirical formula:
It is the simplest formula gives the ratio of atoms of different elements in small whole number
Given data:
Percentage of hydrogen = 13.3%
Percentage of carbon = 52.0%
Percentage of oxygen = 34.7%
Empirical formula = ?
Molecular formula = ?
Solution:
Number of gram atoms of H = 13.3 / 1.01 = 13
.17
Number of gram atoms of O = 34.7 / 16 = 2.17
Number of gram atoms of C = 52.0 / 12 = 4.3
Atomic ratio:
C : H : O
4.3/2.17 : 13.17/2.17 : 2.17/2.17
2 : 6 : 1
C : H : O = 2 : 6 : 1
Empirical formula is C₂H₆O.
Molecular formula:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
n = 46 / 158
The relative molecular mass of compound is not correct.
A variable that is not being directly tested during an experiment should be independent