The equilibrium constant ( K_{eq} ) for the reaction involving nitrogen gas (( N_2 )) depends on the specific reaction being considered. For example, in the formation of nitrogen gas from its elements, ( K_{eq} ) would reflect the concentrations of products and reactants at equilibrium. If you have a specific reaction in mind, please provide it for a more precise answer.
reactants are favored over products in the reaction
I'm taking an awesome chemistry final tomorrow. So, I'm not a massive failure at this: k=mol/liters Kc can only determine by experiment , not by evaluations of equations. so when writting the eq of Kc= [] products /[reactants], do not use units for [], as Kc has no units. Kc, only affected by temperature...
A rate constant
The balanced reaction between ethanol (C2H5OH) and potassium (K) is 2C2H5OH + 2K --> 2C2H5OK + H2. This reaction forms potassium ethoxide and hydrogen gas as products.
reactants are favored over products in the reaction
The equilibrium constant ( K_{eq} ) for the reaction involving nitrogen gas (( N_2 )) depends on the specific reaction being considered. For example, in the formation of nitrogen gas from its elements, ( K_{eq} ) would reflect the concentrations of products and reactants at equilibrium. If you have a specific reaction in mind, please provide it for a more precise answer.
K-2so
reactants are favored over products in the reaction
The equilibrium constant (K eq) for the reaction 2HCl(g) ⇌ H2(g) + Cl2(g) would be [H2][Cl2]/[HCl]^2, where the square brackets indicate the molar concentrations of the respective species at equilibrium.
I suspect it could be " k eq 1" , or "k =1".
Keq = products/reactions
The correct name of K2SO is K-2SO, which stands for "Kay-Tuesso."
Law of mass action (also called 'Law of Guldberg & Waage')Generalized reaction equation (cf. Note): aA + bB pP + qQAccording to the Law of mass action:Equilibrium constant (K>/) and backward (
Not necessarily. The equilibrium constant (K) quantifies the extent of a reaction at equilibrium, but it does not directly correlate to the rate of reaction. A large equilibrium constant indicates that the reaction favors the products at equilibrium, but the rate of the reaction depends on factors such as concentration, temperature, and catalysts.
MgSO4 + KOH --> MgOH + K2SO4 And I want to think that MgOH is a solid
The CV value is the flow rate required to generate 1 psid of pressure loss through the valve. Since pressure drop follows a basic square law the relationship between Cv, flow and pressure loss is as follows: DP = k x Flow^2 eq 1 Where k is a constant that represents the flow shape in the wide open condition. Since Cv is the flow rate at 1 psi of pressure loss then it follows that DP = k x Cv^2 = 1 eq 2 solving for k from eq 2 yields k = 1/Cv^2 eq 3 substituting eq 3 into eq 1 yields DP = (Flow/Cv)^2 Now you have an equation that will tell you the pressure and flow relationship for that particular valve with a particular fully open Cv value. In short, the higher the Cv value the more flow the valve will allow for the same pressure loss or the less pressure loss for the same flow. Good luck