Chemistry Equilibrium

Convert Kc to Kp. Given: Kc=6 Δn = 2 760. °C

Kp = (6) [(0.08206) (1033)]^2 Kp = (6) (7185.61) Kp = 43113 Use the values given to plug in the Kp = Kc(RT)^(Δn) equation. Change 760. °C to Kelvin. Use the given Kc. and plug in the constant .08206.

What is the equilibrium constant (Kc) for the following reaction?: CO(g) + 2H2(g) ⇌ CH3OH(g) Given: Equilibrium Concentrations at 25.0 °C [H2] = 0.25M [CO] = 1.25 M [CH3OH] = 3.2 M

[CH3OH] Kc = -------- [CO] [H2]^2 [3.2] Kc = -------- =40.96 [1.25] [.25]^2 You need to substitute the concentrations for the Kc equation. Then you solve to get the answer of the Kc=40.96.

What is the equilibrium concentration of C? K=2*10^-6 A(aq) + B(s) = C +2D [A]=4.5+10^-4 [B]=16 [C]=x [D]=1.2*10^-2

[C][D]^2 Kc = -------- [A][B] [x][1.2*10^-2]^2 Kc = ---------------- [4.5+10^-4][B] x= 6.258*10^-6 Plug in the Products over reactants. Get rid of B because it is a solid. Plug in the values to get on 6.258*10^-6.

What is the equilibrium constant (Kc) for the following reaction?: H2(g) + I2(g) ⇌ 2HI(g) Given: Equilibrium Concentrations at 25.0 °C [H2] = 0.5 M [I2] = 0.2 M [HI] = 2 M

[HI]^2 Kc = -------- [H2] [I2] [2]^2 Kc = -------- =40 [.5] [.2] You need to substitute the concentrations for the Kc equation. Then you solve to get the answer of the Kc=40.

For the equation 2A + B = 4C, the value of the equilibrium constant, K is 20. What is the value of the equilibrium constant for the reaction below? 8C = 4A + 2B K= ? Given: 2A + B = 4C K=20 a. 0.0025 b. .0010 c. .02540 d. .20000

a. 0.0025 When you multiply the reaction the new K is K^x. X being that amount. In this case it is 1/K^2. So the answer is 1/20^2=0.0025

What is the formula for the equilibrium constant? a. Kc= [C]^c[D]^d/ [A]^a[B]^b b. aA + bB -> cC + dD c. Q= [C]^c[D]^d/ [A]^a[B]^b d. 2SO2(g) + O2 (g) = 2SO3 (g)

a. Kc= [C]^c[D]^d/ [A]^a[B]^b The formula for the equilibrium constant is Products/Reactants. Kc represents equilibrium.

What does it mean when Q < K? a.More reactants b.More products c.Equal amount of reactants and products d.No reactants or products

a.More reactants The reaction will favor the forward reaction to reach equilibrium. In order for this to happen then the denominator must be bigger than the numerator.

Does the equilibrium constant K have units? a.Yes b.No

a.Yes From a mathematical approach the mol⋅L−1 for a product divided by a reactant of the same unit leaves a unitless answer. Most chemistry books leave Kc unitless.

During equilibrium can reactions occur in both the forward and backward directions? a.Yes b.No

a.Yes Reactions during equilibrium can go in the forward or backward reaction because they reach a state of balance. Reactants to products and products to reactants. The concentrations of reactants and products stay constant.

Does heat have an effect on the position of equilibrium? a.Yes b.No

a.Yes The heat causes equilibrium to shift towards the direction of the side that absorbs the heat.

Do catalysts increase the rate of both forwards and reverse reactions? a.Yes b.No

a.Yes They will speed up both the forward and reverse reactions, but they will not have any effect on chemical change.

What is the reaction quotient? a. H2O+CO2->C6H12O6+O2 b. Q= [C]^c[D]^d/ [A]^a[B]^b c. Q=K d. Q>K

b. Q= [C]^c[D]^d/ [A]^a[B]^b The reaction quotient has the same structure of the Equilibrium constant expression. It is Q= to Concentration of Products over the concentration of Reactants.

All types of systems can reach equilibrium? a.True b.False

b.False Open systems cannot reach equilibrium because an open system allows outside resources to affect the reaction or take away one side of the reaction. Like the product's side could be taken away or vice versa. This causes no net change which is not at equilibrium.

What does it mean when Q > K? a.More reactants b.More products c.Equal amount of reactants and products d.No reactants or products

b.More products The reaction will favor the reverse reaction to reach equilibrium. In order for this to happen then the numerator must be bigger than the denominator.

Do catalysts have any effect on the final equilibrium position of the reaction? a.Yes b.No

b.No Catalysts do not affect the final position of equilibrium. They only speed the reaction. Other than that the reaction does not undergo any chemical change from catalysts.

Can volume and pressure mess with equilibrium? a.Yes b.No c.Depends on the situation

c. Depends on the situation The volume and pressure can only mess with equilibrium when the number of moles of gas is not equal to each other on the reactant and product sides of the reaction.

What does it mean when Q = K? a.More reactants b.More products c.Equal amount of reactants and products d.No reactants or products

c.Equal amount of reactants and products The reaction is at equilibrium. The rates of the forward and reverse reactions are the same so the concentrations will not change.

In the equation 2A + B = 4C, the value of the equilibrium constant, K is 20. What is the value of the equilibrium constant for the reaction in the reverse way. a.20 b. .20 c. 2 d. .05

d. .05 When you reverse the reaction k becomes 1/k. So 1/20 is .05.

Find the equilibrium amount if the reaction if the initial amount is .5 mols and the change in amount is -.150?

.350 Mol .350 mol because the equilibrium amount is from the initial plus or minus the change. The .5-.15=.350 mol.

3X(g)↽−−⇀4Y(g)+6Z(g) Set up an ice table for the sample amount of .5 mol.

Any setup of an ice table will do. As long as you know how to make an ice table this is a free point.

What is not included in the equilibrium expression when calculating for Kc? (Select all that apply) a.Pure Solids b.Pure Liquids c.Pure Gases d.Pure Metals

Correct answers are a and b. Solids and pure liquids. Pure solids and pure liquids are not included in the equilibrium constant expression because they are equal to 1. They have no effect on the amount of reactants at equilibrium.