AP Chemistry: Equilibrium

2H2S(g) --> 2 H2(g) + S2(g) A rigid reaction vessel contains H2S(g) in equilibrium with H2(g) and S2(g) at 1200 K, as represented by the equation above. The partial pressures of the gases at equilibrium are shown in the table below. Gas Equilibrium partial pressure (atm) ------------------------------------------------ H2S 1.2 H2 0.21 S2 0.016 Calculate the value of the equilibrium constant, Kp, for the reaction of 1200 K

= (.016)(.21)^2/ (1.2)^2 =.00049

N2(g) + O2(g) --> 2NO(g) K1= 2 x 10 -31 2NO(g) + O2(g) --> 2NO2(g) K2= 5 x 10 12 Given the values of K1 and K2 for the reactions represented above, what is the value of K3 for the following reaction? N2(g) + 2O2(g) --> 2NO2(g) K3 = ?

K3 = K1K2 K3 = (2 x 10 -31)(5 x 10 12) = (2 x 5) (10 -31 + 12) = 10 x 10 -19 = 1 x 10 -18

AgBr(s) --> Ag+ (aq) + Br - (aq) K1= 5 x 10 -13 Ag+(aq) + 2NH3(aq) --> Ag(NH3)2+(aq) K2= 2 x 10 7 Given the values of K1 and K2, for the reactions represented above, what is the value of K3 for the following reaction? AgBr(s) + 2NH3(aq) --> Ag(NH3)2+(aq) + Br-(aq) K3 = ?

K3​​=(5×10−13) (2×107) =(5×2)(10−13+7) =10×10−6 =1×10−5​

At 940 K, the equilibrium constant for the following reaction has a value of 16. 2SO2(g) + O2(g) --> 2SO3(g) What is the value of the equilibrium constant for the reaction represented below at 940 K? SO3(g) --> SO2(g) + 1/2 O2(g)

Keq = 1/16 Take the 1/2 to the 16 and it would be 0.25

H2(g) + I2(s) --> 2HI(g) Kp= 0.80 at 315 K The reaction between hydrogen gas and solid iodine is represented above. In an experiment, a student places samples of H2(g) and I2(s) into a previously evacuated, rigid vessel at 315 K. The system is allowed to reach equilibrium, at which point some I2(s) still remains in the vessel. If the partial pressure of HI(g) at equilibrium is 2.0 atm, what is the partial pressure of H2 (g)

Kp= (PHI)^2/(PH2) PH2=(2.0)^2/ .8 = 5

COCl2 (g) --> CO (g) + Cl2 (g) A student is investigating the decomposition reaction represented above. The student begins with a 0.500 mol sample of pure COCl2​(g) in a rigid 1.00 L container at 910 K. When equilibrium is reached, 37.6 percent of the COCl2(g) sample has decomposed. Using molar concentrations, calculate the value of the equilibrium constant, Kc for the reaction at 910 K.

Rice table COCL decomposed is (.376)(.500 M) = .188 M = .500 - x = 0.312 M CO and Cl2 x = .188 M Now plug it in (.188)^2 /.312 = .113

I2 (g) + Br 2 (g) --> 2IBr(g) Kp = 73.3 at 475 K A rigid vessel initially contains I2(g) and Br2(g) each at a partial pressure of 2.00 atm at 475 K. The system comes to equilibrium according to the equation above. Calculate the partial pressure of IBr(g) in the vessel at equilibrium at 475 K

Rice table Expression and plug in = (2x)^2 / (2.00 - x)^2 Square root 73.3 = 2x / 2.00 - x = 1.62 plug in 1.62 to 2x = 3.24

2NOBr(g) --> 2 NO(g) + Br2(g) A rigid reaction vessel contains NOBr(g) in equilibrium with NO(g) and Br2(g) at 450 K, as represented by the equation above. The partial pressures of the gases at equilibrium are shown in the table below. Gas Equilibrium partial pressure (atm) ------------------------------------------------ NOBr 0.32 NO 0.60 Br2 0.98 Calculate the value of the equilibrium constant, Kp, for the reaction of 450 K

Set up the Expression then plug it in Kp= (PNO)^2(PBr2)/(PNOBr)^2 (.60)^2(.98) / (.32)^2 = 3.4

X2 (g) + Y2 (g) --> 2XY(g) Kc=.15 The equation above represents the synthesis of a compound XY(g). The diagram below depicts a mixture of the three species in a 1.0 L rigid vessel at a certain temperature. Each molecule in the diagram represents 1.0 mol of compound. In which direction will the reaction proceed to reach equilibrium at this temperature?

So, the reaction will proceed toward the reactants.

I2 (g) --> 2I(g) The dissociation of I2(g) is represented by the equation above. The value of Keq for the reaction at 950 K is 9.1 x 10 -4 From this information, which of the following can be concluded about the reaction at 950 K?

The reactant is favored over the product at equilibrium. When a reaction has a small equilibrium constant Keq < 1, the reactants are favored over the products at equilibrium.

X(g) + Y(g) --> 2Z(g) + Q(g) A 6.00 mol sample of X(g) and a 6.00 mol sample of Y(g) are combined in a rigid, evacuated 1.00 L container at a particular temperature, and the reaction represented above occurs. When the mixture reaches equilibrium, 5.00 mol of Q(g) is present in the container. What is the value of the equilibrium constant, Kc, for the reaction at this temperature?

To find the value of Kc​, for the reaction, let's first use the balanced equation to write out the equilibrium constant expression: Kc= Z^2 x Q/ X x Y Plugin the values (10.0)^2 (5.00) / (1.00) (1.00) = 500.

X(g) + 2 Q(g) --> 4Z(g) + R(g) Equimolar samples of X(g) and Q(g) are pumped into an evacuated, rigid container and allowed to reach equilibrium according to the equation above. Which of the following must be true at equilibrium?

[Z] must be greater than [R]

Reaction Keq 2BrCl(g) --> Br2(g) + Cl2(g) K1 2IBr(g) --> I2(g) + Br2(g) K2 Based on the information above, which of the following expressions corresponds to Keq for the reaction represented below? 2BrCl(g) --> 2IBr(g) + Cl2(g)

first equation is consistent = K1 second equation adds 2 mol of IBr = 1/K2 = K1/K2

2NO(g) + Cl2 (g) --> 2NOCl(g) Kc= 8.6 x 10 6 at 350 K NO(g) reacts with Cl2(g) to form NOCl(g) according to the equation above. A sample of each gas is injected into a rigid, evacuated container at 350 K. The initial concentrations of the gases are listed in the table below. Gas Initial concentration (M) NO .10 Cl2 .10 NOCl .80 Which of the following best predicts how the total pressure in the vessel will change as the system moves toward equilibrium at 350 K?

the total pressure in the vessel will also decrease.

X(g) + Q(g) --> R(g) + 2Z(g) A 8.00 mol sample of X(g) and a 8.00 mol sample of Q(g) are combined in a rigid, evacuated 1.00 L container at a particular temperature, and the reaction represented above occurs. When the mixture reaches equilibrium, 4.00 mol of R(g) is present in the container. What is the value of the equilibrium constant, Kc, for the reaction at this temperature?

Kc = (4.0)(x)^2/(8)(8) = 16

SbCl5(g) --> SbCl3 (g) + Cl2(g) The decomposition of SbCl5(g) is represented by the equation above. A pure sample of SbCl5(g) is introduced into a rigid, evacuated flask at an initial pressure of 1.50 atm and the temperature is held constant until equilibrium is established. The total pressure in the flask at equilibrium is 1.90 atm. Which of the following is true about the equilibrium constant, Kp, for the reaction?

Rice table Ptotal​ = (1.50−x) + x + x = 1.50 + x = 1.90 atm Solving for x, we get x = 1.90−1.50=0.40 atm Now, using the expressions involving x from our ICE table, let's determine the partial pressure of each gas at equilibrium: PSbCl5​​ = 1.5 - x = 1.10 atm PSbCl3​​ = PCl2​​​ = x = 0.40 atm​ Finally, let's substitute these values into the equilibrium constant expression for Kp: Kp= (PSbCl3) (PCl2)/ (PSbCl5) = (.40)^2 / 1.10 < 1 ------------------------------------- 0 < Kp < 1

Br2(g) + Cl2(g) --> 2BrCl(g) The reaction system represented above is at equilibrium at 298 K. The concentrations of Br2(g) and Cl2(g) under these conditions are 0.015 M and 0.050 M, respectively. Given that the equilibrium constant, Kc, for the reaction is 7.8 at 298 K, determine the concentration of BrCl(g) in the equilibrium mixture.

Square root (7.8)(.015)(.050) = .076