Kc = concentrations of product / concentrations of reactant
Kc = [Br₂] [Cl₂]₃ / [BrCl₃]₂
What is the equilibrium constant?
The relationship between a reaction's products and reactants with regard to a certain unit is expressed by the equilibrium constant(K) This article introduces the mathematics needed to determine the partial pressure equilibrium constant as well as how to formulate expressions for equilibrium constants. By allowing a single reaction to reach equilibrium and then measuring the concentrations of each chemical participating in that reaction, one can determine the numerical value of an equilibrium constant. it is the ratio of product concentrations to reactant concentrations. The equilibrium constant for a given reaction is unaffected by the initial concentrations because the concentrations are measured at equilibrium.
To learn more about the equilibrium constant, visit:
brainly.com/question/19340344
#SPJ4
Okay give me a few minutes and I will send u a screen shot of my answers.
(2) Adding more O2(g) would shift the equilibrium to the right because a higher concentration of oxygen is offered than its initial position, therefore more products have to be yielded to maintain equilibrium.
Hopes this helps:
Answer: So the pH of the solution is 4.
Answer:
ΔH°f(C₈H₁₈(g)) = -210.9 kJ/mol
Explanation:
Let's consider the combustion of C₈H₁₈.
C₈H₁₈(g) + 25/2 O₂(g) ⟶ 8 CO₂(g) + 9 H₂O(g) ΔH°rxn = − 5113.3 kJ
We can calculate the standard enthalpy of formation of C₈H₁₈(g) using the following expression.
ΔH°rxn = 8 mol × ΔH°f(CO₂(g)) + 9 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C₈H₁₈(g)) - 25/2 mol × ΔH°f(O₂(g))
1 mol × ΔH°f(C₈H₁₈(g)) = 8 mol × ΔH°f(CO₂(g)) + 9 mol × ΔH°f(H₂O(g)) - 25/2 mol × ΔH°f(O₂(g)) - ΔH°rxn
1 mol × ΔH°f(C₈H₁₈(g)) = 8 mol × (-393.5 kJ/mol) + 9 mol × (-241.8 kJ/mol) - 25/2 mol × 0 kJ/mol - (− 5113.3 kJ)
ΔH°f(C₈H₁₈(g)) = -210.9 kJ/mol