Chapter 15 A
At 1000 K, Kp=1.85 for the reaction SO2(g)+1/2O2(g)←→SO3(g) What is the value of Kp for the reaction SO3(g)←→SO2(g)+1/2O2(g)
* ASKS for the REVERSE Kp * Kp = Pso3 / (Pso2) (Po2^1/2) = 1.85 Kp (r) = (Pso2) (Po2^1/2) / Pso3 = 1 / 1.85 = 0.541
What does a small equilibrium constant (k) indicate?
* k << 1 (small k): equilibrium lies to the left * This equilibrium indicates that the equilibrium mixture contains mostly reactants
What does a large equilibrium constant (k) indicate?
*k >>1 (large k): equilibrium lies to the right * This equilibrium indicates that the equilibrium mixture contains mostly products
Which of the following are equal at equilibrium? 1) Forward and reverse reaction rates 2) Reactant and product concentrations
1) Forward and reverse reaction rates
Which of the following variables are equal when the N2O4(g)⇌2 NO2(g) reaction reaches equilibrium: 1) Kf & Kr 2) Concentration [N2O4] & [NO2] 3) Forward and Reverse Reaction Rates
3) Forward and Reverse reaction rates
Snapshots of two hypothetical reactions, A(g)+B(g)⇌AB(g) (4 AB product molecules) and X(g)+Y(g)⇌XY(g) (3 XY product molecules) Which has the larger equilibrium constant?
A(g)+B(g)⇌AB(g); (4 AB product molecules) The larger amount of products present at equilibrium the larger the equilibrium constant
The following diagram represents a reaction shown going to completion. Each molecule in the diagram represents 0.1 mol, and the volume of the box is 1.0 L 1) 5 A2 molecules & 5 B molecules 2) 4 A2B molecules, 1 A2 and 1 B molecule Letting A = red spheres and B = blue spheres, write a balanced equation for the reaction.
A2 + B --> A2B
The equilibrium constant for the dissociation of molecular chlorine, Cl2(g)⇌2Cl(g), at 200 ∘C is Kc=0.0168. Which species predominates at equilibrium, Cl2 or Cl?
Cl2 (reactant) predominates We know this because the k value is much less than 1, which also means the denominator of the k expression is much larger than the numerator
Two different proteins X and Y are dissolved in aqueous solution at 37 ∘C. The proteins bind in a 1:1 ratio to form XY. A solution that is initially 1.00 mM in each protein is allowed to reach equilibrium. At equilibrium, 0.18 mM of free X and 0.18 mM of free Y remain. What is Kc for the reaction?
Equation = X (aq) + Y (aq) <=> XY (aq) *Set up ICE table* X (aq) + Y (aq) <=> XY (aq) (I) 1.0 mM 1.0 mM 0 (C) (E) 0.18 mM 0.18 mM ? - Find Changes - Solve Kc Kc = [XY] / [X] [Y] = [0.82 x 10^-3] / [ 0.18 x 10^-3] ]0.18 x 10^-3] = 2.5 x 10^4
The equilibrium constant for the reaction 2NO(g)+Br2⇌2NOBr(g) is Kc=1.3×10−2 at certain temperature. Calculate Kc for NOBr(g)⇌NO(g)+( 1/2Br2)(g).
Kc (reverse) = ([NO] [Br2]^1/2) / [NOBr2] = Kc (reverse)^1/2 = (76.92)^1/2 = 8.8
The equilibrium constant for the reaction 2NO(g)+Br2⇌2NOBr(g) is Kc=1.3×10−2 at certain temperature. Calculate Kc for 2NOBr(g)⇌2NO(g)+Br2(g).
Kc (reverse) = reciprocal of Kc (forward) Kc (forward) = ([NOBr2]^2) / ([NO]^2 [Br2]) = 1.3x10^-2 *Kc (reverse) = ([NO]^2 [Br2]) / [NOBr2]^2 = 1 / (1.3x10^-2) = 76.92*
At 1000 K, Kp=1.85 for the reaction SO2(g)+1/2O2(g)←→SO3(g) What is the value of Kp for the reaction 2SO2(g)+O2(g)←→2SO3(g)
Kc = Pso3^2 / Pso2^2 x Po2 = (1.85)^2 = 3.42
If Kc = 4.4×10−2 for PCl3(g)+Cl2(g)⇌PCl5(g) at 490 K What is the value of Kp for this reaction at this temperature?
Kp = Kc (RT) ^ Δn Kp = (4.4x10^-2) (R x 490)^-1 Kp = 1.1 x 10^-3
The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures of 9.80×10−2 atm , 0.175 atm , and 0.27 atm for NO, Cl2, and NOCl, respectively. If the vessel has a volume of 4.00 L , calculate Kc at this temperature.
Kp = Kc (RT)^-1 Kc = Kp (RT) = (43) (.08206) (500) = 1.8 x 10^3
Calculate the value of Kp at 25 ∘C, where Kc = 40 A2 + B --> A2B
Kp = Kc (RT)^Δn Kp = (40) x (0.08206 L-atm/mol-k x 298 K)^-1
The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures of 9.80×10−2 atm , 0.175 atm , and 0.27 atm for NO, Cl2, and NOCl, respectively. Calculate Kp for this reaction at 500.0 K.
Kp = PNOCl^2 / NO^2 x Cl2 = (0.270^2 / (9.8x10^-2)^2 (0.175) = 43
The more _________ present at equilibrium the ___________ the equilibrium constant
Product, larger
Calculate the value of Kc A2 + B --> A2B
Since the reaction is shown going to completion, the rightmost box will most accurately depict the equilibrium concentrations. Using the molar quantities and volume given, and the proper equation from Part B, you can calculate the equilibrium constant. Kc =[A2B] / [A2][B] = 0.4 M / [0.1 M][0.1 M] Kc = 40
The equilibrium constant for the dissociation of molecular chlorine, Cl2(g)⇌2Cl(g), at 200 ∘C is Kc=0.0168. Assuming both forward and reverse reactions are elementary processes, which reaction has the larger rate constant: the forward or the reverse reaction?
The reverse reaction has the greater rate constant. K c = k f / k r If Kc is small, kr is larger than kf
Choose the equilibrium constant expression for the reaction. A2 + B --> A2B
k = [A2B] / [A2] [B] K = ratio of concentration of products over reactants Raised to the power of their stoichiometric coefficients respectively
The equilibrium constant for the reaction 2NO(g)+Br2⇌2NOBr(g) is Kc=1.3×10−2 at certain temperature. At this temperature does the equilibrium favor NO and Br2, or does it favor NOBr?
the equilibrium favors NO & Br2 We know this because the Kc value is less than 1, which indicates the reaction favors the reactants
Assuming that all of the molecules are in the gas phase, calculate Δn, the change in the number of gas molecules that accompanies the reaction. A2 + B --> A2B
Δn = 1−2=−1 Δn represents the difference in the number of moles of gas present in the reactants versus the products. The balanced chemical equation shows two moles of gas on the reactant side (A2 and B) and one mole of gas on the product side (A2B). The difference between these two values will give Δn. Δn=(moles of gaseous product)−(moles of gaseous reactant)