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equilibrium constant
Constant TermDefinition of Constant TermThe Constant Term in an expression or equation has a fixed value and does not contain variables.Examples of Constant TermThe constant term in the equation 6x2 - 3x + 5 is 5.Solved Example on Constant TermChoose the value of the constant term in the expression,- 12x2 + 5x +1.Choices:A. 5B. 1C. 0D. - 12Correct Answer: BSolution:Step 1: - 12x2 + 5x +1. [Original expression.]Step 2: The Constant Term in an algebraic expression or equation has a fixed value and does not contain variables.Step 3: Therefore, 1 is the constant term in the expression, - 12x2 + 5x +1.There You Go Baby ;]
The constant. For instance, if you had 2x +5, +5 would be your constant, because no matter what number you substitute in for x, the last term of the expression will be +5. It is independent of the x and y value.
The question is about an oxymoronic expression. A constant cannot be a variable and a variable cannot be a constant!
Kc is the equilibrium constant.
To find the equilibrium concentration of NO, first calculate the equilibrium constant expression using the given concentrations of O2 and N2. Then, rearrange the equilibrium constant expression to solve for the concentration of NO. Finally, substitute the values of O2 and N2 concentrations into the rearranged expression to find the equilibrium concentration of NO.
The correct form for the equilibrium constant expression for this reaction is Kc = [HF]^2 / ([H2] * [F2]), where the square brackets denote molar concentrations of each species at equilibrium.
To write an equilibrium constant expression using a balanced chemical equation, you need to identify the reactants and products involved in the equilibrium and write the expression as a ratio of the products raised to their stoichiometric coefficients divided by the reactants raised to their stoichiometric coefficients. The general format is [products]/[reactants]. The coefficients from the balanced equation become the exponents in the expression.
In the Ksp expression, only the concentration of the dissolved ions is included because the solid is considered to be in equilibrium with the ions in solution. Therefore, its concentration remains constant and is not included in the expression. Including the solid in the Ksp expression would not affect the equilibrium constant value.
The equilibrium constant Kc is a ratio of the concentrations of products to reactants, and these units cancel out to give a unitless value. It represents the equilibrium position of a reaction, regardless of the units used for concentration.
A large value of Keq (equilibrium constant) indicates that the reaction favors the formation of products at equilibrium. This means that the forward reaction is favored and the concentration of products is higher than that of reactants in the equilibrium state.
The expression for the force constant (k) in Hooke's Law is given by the equation F = kx, where F is the force applied, k is the force constant, and x is the displacement from equilibrium. The force constant is a measure of the stiffness of a spring or a bond.
The equilibrium constant expression for H2SO3 is K = [H+]^2[HSO3-]/[H2SO3].
Given the equilibrium constant (Kc) is 0.625 and the concentrations of O2 and H2O at equilibrium are 0.40 and 0.20 respectively, you can use the equilibrium expression Kc = [H2O2] / ([O2] * [H2O]) to solve for the equilibrium concentration of H2O2. Plugging in the values, you can calculate the concentration of H2O2 at equilibrium.
The base dissociation constant (Kb) for a weak base is the equilibrium constant for the reaction of the base with water to produce hydroxide ions. In this case, the expression for Kb would be [OH-][BH]/[B].
The acid dissociation constant (Ka) for an acid HX at equilibrium is the ratio of the concentrations of the products (H+ and X-) to the undissociated acid (HX) in the equilibrium expression for the dissociation reaction. It is a measure of the strength of the acid, with higher Ka values indicating a stronger acid.