CHEM - 2

Ace your homework & exams now with Quizwiz!

Which of the following is a (conjugate base, conjugate acid) pair?

(CN-, HCN)

State whether HC2H3O2 is a strong or weak acid.

Weak Correct.

State whether HClO is a strong or weak acid.

Weak Correct.

Which of the following describes the effect of adding reactants to a reaction mixture at equilibrium? (Q is the reaction quotient, Keq is the equilibrium constant)

decrease in Q; reaction shifts to the right until Q=Keq again

Which of the following is a (conjugate acid, conjugate base) pair?

(H2O, OH-)

Consider a system at 25oC where the following reaction is occurring with Kp = 7.45: A(g) + B(g) ⇌ C(g) + D(g) What is the partial pressure of B at equilibrium if the initial partial pressures of the products at 25oC are: Partial pressure of C, atm0.828 Partial pressure of D, atm0.828

0.222 atm Correct. Kp=7.45= (0.828-x)^2 / x^2 SQUARE ROOT 2.73 = .828 - x / x 2.73x + x = 3.73 x = .828 /3.73 = .222 atm

Consider the hypothetical reaction: A(s) = B(aq) + 2 C(aq) If at a given instant, the concentrations of B and C are 2.28x10^-8 and 1.01x10^-7 mol/L, the value of the reaction quotient (Q) is

2.32x10^-22 Correct. Q = [B][C]^2 / 1 = (2.28x10-8)(1.01x10-7)^2 = 2.32x10^-22

True or False. Once a reaction reaches equilibrium, there is no further transformation of reactant molecules into product molecules.

False

State whether C6H5NH2 is a strong or weak base.

Weak Correct. The strong bases are LiOH, NaOH, KOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2.

Consider the following fictitious reaction at equilibrium: A(g) + B(g) ⇌ C(g) ΔH = -80 kJ Which of the following stresses would not shift the equilibrium?

adding a catalyst

Which of the following describes the effect of cooling a reaction mixture at equilibrium, if the forward reaction is endothermic? (Keq is the equilibrium constant)

decrease in Keq; reaction shifts to the left

Consider the following reaction: 2 NH3(g) ⇌ N2(g) + 3 H2(g) If a vessel contains only 0.176 atm of pure NH3 and 0.110 atm N2 initially and, at equilibrium the pressure of H2 was found to be 3x atm, we can calculate the equilibrium constant Kp using

(0.110 + x)(3x)^3 / (0.176 - 2x)^2 Correct. The initial pressures, in atm, are: P_NH3 = 0.176 P_N2 = 0.110 P_H2 = 0 From the stoichiometry, assuming constant volume, the equilibrium pressures are: P_NH3 = 0.176 - 2x P_N2 = 0.110 + x P_H2 = 3x The equilibrium constant is: KP = (P_N2)(P_H2)3/(P_NH3)2 = (0.110+x (3x)3/(0.176-2x)2

Consider the following reaction: 2 NH3(g) ⇌ N2(g) + 3 H2(g) If a vessel contains only 0.157 atm of pure NH3 and 0.124 atm N2 initially and, at equilibrium the pressure of H2 was found to be 3x atm, we can calculate the equilibrium constant Kp using

(0.124 + x)(3x)^3 / (0.157 - 2x)^2 Correct. The initial pressures, in atm, are: P_NH3 = 0.157 P_N2 = 0.124 P_H2 = 0 From the stoichiometry, assuming constant volume, the equilibrium pressures are: P_NH3 = 0.157 - 2x P_N2 = 0.124 + x P_H2 = 3x The equilibrium constant is: KP = (P_N2)(P_H2)3/(P_NH3)2 = (0.124+x)(3x)3/(0.157-2x)2

Consider the following reaction: 2 NH3(g) ⇌ N2(g) + 3 H2(g) If a vessel contains only 0.213 atm of pure NH3 initially and, at equilibrium the pressure of N2 was found to be x atm, we can calculate the equilibrium constant Kp using

(x)(3x)^3 / (0.213 - 2x)^2 Correct. The initial pressures, in atm, are: P_NH3 = 0.213 P_N2 = 0 P_H2 = 0 From the stoichiometry, assuming constant volume, the equilibrium pressures are: P_NH3 = 0.213 - 2x P_N2 = x P_H2 = 3x The equilibrium constant is: KP = (P_N2)(P_H2)3/(P_NH3)2 = (x)(3x)3/(0.213-2x)2

Consider the following reaction: 2 NH3(g) ⇌ N2(g) + 3 H2(g) If a vessel contains only 0.316 atm of pure NH3 initially and, at equilibrium the pressure of N2 was found to be x atm, we can calculate the equilibrium constant Kp using

(x)(3x)^3 / (0.316 - 2x)^2 Correct. The initial pressures, in atm, are: P_NH3 = 0.316 P_N2 = 0 P_H2 = 0 From the stoichiometry, assuming constant volume, the equilibrium pressures are: P_NH3 = 0.316 - 2x P_N2 = x P_H2 = 3x The equilibrium constant is:KP = (P_N2)(P_H2)3/(P_NH3)2 = (x)(3x)3/(0.316-2x)2

Consider the following reaction: 2 NH3(g) = N2(g) + 3 H2(g) If a vessel contains only 0.429 atm of pure NH3 initially and, at equilibrium the pressure of N2 was found to be x atm, we can calculate the equilibrium constant Kp using

(x)(3x)^3 / (0.429 - 2x)^2 Correct. The initial pressures, in atm, are: P_NH3 = 0.429 P_N2 = 0 P_H2 = 0 From the stoichiometry, assuming constant volume, the equilibrium pressures are: P_NH3 = 0.429 - 2x P_N2 = x P_H2 = 3x The equilibrium constant is: KP = (P_N2)(P_H2)3/(P_NH3)2 = (x)(3x)3/(0.429-2x)2

Consider a system where the following reaction is occurring: 3 A(aq) + 4 B(l) ⇌ 3 C(aq) + 4 D(aq) The equilibrium molar concentrations are: [A]0.308 [C]0.441 [D]0.321

.0312 Correct. Kc = C^3 * D^4 / A^3 Kc = .441^3 * .321^4 / .308^3 = .0312

Suppose the reaction quotient (Q) for a reaction mixture at a particular instant is 2.05 if the reaction is written as 2 NH3 ⇌ N2 + 3 H2 Calculate the reaction quotient if the reaction is written as 5 N2 + 15 H2 ⇌ 10 NH3.

0.0276 Correct. We can see that we multiplied all the coefficients by 5. Therefore, the POWERS in the expression for Q will all be multiplied by 5. However, we have also flipped the equation. This means that we also need to filp the numerator and denominator in the expression for Q. Therefore: Q = Reciprocal of [(original Q)5] = 1/{(2.05)5} = 0.0276

Consider a system where the following reaction is occurring: 4 A(g) ⇌ B(g) + 3 C(g) The equilibrium partial pressures are: Partial pressure of A, atm 0.603 Partial pressure of B, atm0.467 Partial pressure of C, atm 0.221 Calculate the value of the equilibrium constant, Kp.

0.0381 Correct. Kp = (PB) * (PC)^3 / (PA)^4 Kp = (.603) * (.467)^3 / (.788)^4 = 0.0381

At a certain temperature, the equilibrium constants are known for the chemical equations 4 A(g) ⇌ 4 B(g) + 4 C(g) K = 3.08x10^-2 2 D(g) ⇌ 2 E(g) + 2 B(g) K = 2.81x10^-2 At the same temperature, calculate the equilibrium constant for 2 C(g) + 2 D(g) ⇌ 2 E(g) + 2 A(g)

0.160 Correct. 2 D(g) ⇌ 2 E(g) + 2 B(g) K = 2.81x10^-2 2 B(g) + 2 C(g) ⇌ 2 A(g) K' = 1 / (3.08x10-2)^1/2 = 5.70 K = 5.70 * 2.81x10^-2 = 0.160

At a certain temperature, the equilibrium constants are known for the chemical equations 4 A(g) ⇌ 6 B(g) + 4 C(g) K = 5.61x10^-2 3 D(g) ⇌ 3 E(g) + 3 B(g) K = 5.61x10^-2 At the same temperature, calculate the equilibrium constant for 2 C(g) + 3 D(g) ⇌ 3 E(g) + 2 A(g)

0.237 Correct. 3 D(g) ⇌ 3 E(g) + 3 B(g) K = 5.61x10-2 3 B(g) + 2 C(g) ⇌ 2 A(g) K' = 1 / (5.61x10-2)^1/2 = 4.22 K = 4.22 * 5.61x10^-2 = 0.237

Consider a system where the following reaction is occurring: 4 A(g) ⇌ B(g) + 2 C(g) The equilibrium partial pressures are: Partial pressure of A, atm 0.788 Partial pressure of B, atm 0.284 Partial pressure of C, atm 0.579 Calculate the value of the equilibrium constant, Kp.

0.247 Correct. Kp = (PB) * (PC)^2 / (PA)^4 Kp = (.284) * (.579)^2 / (.788)^4 = 0.247

Consider a system at 25oC where the following reaction is occurring with Kp = 3.91: A(g) + B(g) ⇌ C(g) + D(g) What is the partial pressure of B at equilibrium if the initial partial pressures of the products at 25oC are: Partial pressure of C, atm0.828 Partial pressure of D, atm0.775

0.269 atm Correct. FIRST: A = 3.91 - 1 = 2.91 SECOND: B = .828 + .775 = 1.60 THIRD: C = .828 * .775 = .642 = -.642 PLUG INTO: x = -b ± (b^2 -4ac)^1/2 / 2a x = -1.60 ± [(1.60)^2 -4(2.91)(-0.642)]^1/2 / 2(2.91) = 0.269 atm

Consider the following representation of a equilibrium system where the reaction 3 A(g) ⇌ 2 B(g) is involved. Assuming each red circle represents 0.00786 atm of A and each blue circle represents 0.00203 atm of B, what is the value of Kp?

0.353 Correct. From the figure, one can see that there are 6 red circles representing A and 3 blue circles representing B. This means that the partial pressures of A and B are PA = 6 * 0.00786 = 0.0472 atm PB = 3 * 0.00203 = 0.00609 atm Kp = (Pb)^2 / (Pa)^3 Kp = (0.00609)^2 / (0.0472)^3 = 0.353

Consider a system at 25oC where the following reaction is occurring with Kp = 0.802: A(g) + B(g) ⇌ C(g) + D(g) What is the partial pressure of C at equilibrium if the initial partial pressures of the reactants at 25oC are:

0.421 atm Correct. Kp=0.802= (PC) * (PD) / (PA) * (PB) Kp=(x)(x) / (0.891 - x)(0.891 - x) Kp = x^2 / (0.891 - x)^2 Since this is a perfect square, the arithmetric involved can be simplifed by taking the square root of both sides giving, 0.896= x/0.891 - x .896*.891 = .798 .896*x=.896x (add 1) x = .798 / 1.896x = 0.421 atm.

Consider a system at 25oC where the following reaction is occurring with Kp = 8.23: A(g) + B(g) ⇌ C(g) + D(g) What is the partial pressure of C at equilibrium if the initial partial pressures of the reactants at 25oC are: Partial pressure of A, atm0.704 Partial pressure of B, atm0.704

0.522 atm Correct. Kp=8.23= (PC) * (PD) / (PA) * (PB) Kp=(x)(x) / (0.704 - x)(0.704 - x) Kp = x^2 / (0.704 - x)^2 Since this is a perfect square, the arithmetric involved can be simplifed by taking the square root of both sides giving, 2.87= x/0.704 - x 2.87*.704 = 2.02 2.87*x=2.87x (add 1) x = 2.02 / 3.87 x = 0.522 atm.

Consider a system where the following reaction occurs at constant volume and temperature: A(g) ⇌ 2 B(g) + C(g) The initial partial pressures are: Partial pressure of A, atm 0.492 Partial pressure of B, atm 0.360 Partial pressure of C, atm 0.499 If the partial pressure of B at equilibrium is 0.577 atm, calculate the value of the equilibrium constant, Kp.

0.526 Correct. FIRST: .577-.360 = .217 SECOND: .217/2 = .108 THIRD: 0.492 atm 0.360 atm 0.499 atm -0.108 atm + 0.217 atm + 0.108 atm LAST: Kp = (B)^2 (C) / (A) Kp = .577^2 * .607 / .384 = .526

Consider a system where the following reaction occurs at constant volume and temperature: 2 A(g) ⇌ B(g) + C(g) The initial partial pressures are: Partial pressure of A, atm 0.246 Partial pressure of B, atm 0.520 Partial pressure of C, atm 0.523 If the partial pressure of A at equilibrium is 0.502 atm, calculate the value of the equilibrium constant, Kp.

0.614 Correct. FIRST: .502-.246 = .256 SECOND: .256/2 = .128 THIRD: 0.246 atm 0.520 atm 0.523 atm +0.256 atm - 0.128 atm - 0.128 atm LAST: Kp = (B) (C) / (A)^2 Kp = .392 * .395 / .502^2 = .614

Consider a system at 25oC where the following reaction is occurring with Kp = 7.04: A(g) + B(g) ⇌ C(g) + D(g) What is the partial pressure of C at equilibrium if the initial partial pressures of the products at 25oC are: Partial pressure of C, atm0.906 Partial pressure of D, atm0.906

0.658 atm Correct. Kp=7.04= (0.906-x)^2 / x^2 SQUARE ROOT 2.65 = .906 - x / x 2.65x + x = 3.65 x = .906 /3.65 = .248 atm WE ARE LOOKING FOR C NOT B .906 - .248 = .658 atm

Consider a system at 25oC where the following reaction is occurring with Kp = 21.5: A(g) + B(g) ⇌ C(g) + D(g) What is the partial pressure of C at equilibrium if the initial partial pressures of the products at 25oC are: Partial pressure of C, atm0.906 Partial pressure of D, atm0.833

0.752 atm Correct. FIRST: A = 21.5 - 1 = 20.5 SECOND: B = .906 + .833 = 1.74 THIRD: C = .906 * .833 = .755 = -.755 PLUG INTO: x = -b ± (b^2 -4ac)^1/2 / 2a x = -1.74 ± [(1.74)^2 -4(20.5)(-0.755)]^1/2 / 2(20.5) = 0.154 atm LOOKING FOR C: 0.906 - 0.154 atm = 0.752

Consider a system at 25oC where the following reaction is occurring with Kp = 0.807: 2 A(g) ⇌ 3B(g) + C(g) At a given instant, the partial pressures at 25oC are: Partial pressure of A, atm0.340 Partial pressure of B, atm0.547 Partial pressure of C, atm0.570 Is the the mixture at equilibrium? If not, which way will the system shift to establish equilibrium? Respond by typing 1, 2, or 3 where 1- the mixture is at equilibrium already 2 - the mixture is not at equilibrium; it will shift right until equilibrium is established 3 - the mixture is not at equilibrium; it will shift left until equilibrium is established

1 Correct. 2 A(g) ⇌ 3B(g) + C(g) Q= ((0.547)^3 * (0.570) / (0.340)^2 = 0.807 Comparing Q and K (Q = 0.807) = (K = 0.807) The answer is 1. The system is already at equilibrium, and nothing happens.

Consider a system at 25oC where the following reaction is occurring with Kp = 2.84: 2 A(g) ⇌ 3B(g) + C(g) At a given instant, the partial pressures at 25oC are: Partial pressure of A, atm0.228 Partial pressure of B, atm0.658 Partial pressure of C, atm0.518 Is the the mixture at equilibrium? If not, which way will the system shift to establish equilibrium? Respond by typing 1, 2, or 3 where 1- the mixture is at equilibrium already 2 - the mixture is not at equilibrium; it will shift right until equilibrium is established 3 - the mixture is not at equilibrium; it will shift left until equilibrium is established

1 Correct. 2 A(g) ⇌ 3B(g) + C(g) Q= ((0.658)^3 * (0.518) / (0.228)^2 = 2.84

At a certain temperature, the equilibrium constant for the chemical equation A(g) + 3B(g) ⇌ C(g) + 2 D(g) is 9.88x10^-2. At the same temperature, calculate the equilibrium constant for 2C(g) + 4D(g) ⇌ 2A(g) + 6B(g)

1.02x10^2 Correct. K = 1/Kn = 1/(9.88x10^-2)^2 = 1.02x10^2

What is the [H3O+] in a solution with a pH of 1.99?

1.0E-2 Correct. [H3O+] = 10^-1.99 = 1.0x10-2 M

What is the concentration of OH- in a 1.12x10^-2 M aqueous solution of NaOH?

1.12E-2 Correct. 1.12x10^-2 M NaOH * 1 mol OH- / 1 mol NaOH = 1.12x10^-2

What is the concentration of H3O+ in a 1.15x10-2 M aqueous solution of HClO4?

1.15E-2 Correct. [H3O+] = [HClO4] = 1.15x10^-2 M .

At a certain temperature, the equilibrium constant for the chemical equation 2A(g) + 3B(g) ⇌ 2C(g) + D(g) is 1.05x10-3. At the same temperature, calculate the equilibrium constant for 6A(g) + 9B(g) ⇌ 6C(g) + 3D(g)

1.16x10^-9 (1.16E-9) Correct. K = Kn = (1.05x10^-3)^3 = 1.16x10^-9

What is the [OH-] in a solution where [H3O+] is 8.32x10^-6?

1.20E-9 Correct. [OH-] = 1.00 x 10^-14 / 8.32x10^-6 = 1.20x10^-9 M

Consider a system at 25oC where the following reaction is occurring with Kp = 0.773: 2 A(g) ⇌ B(g) + C(g) At a given instant, the partial pressures at 25oC are the following: Partial pressure of A, atm1.07 Partial pressure of B, atm1.73 Partial pressure of C, atm1.73

1.44 atm Correct. Kp=.773= (1.73 + x)^2 / (1.07 - 2x)^2 SQUARE ROOT .879 = 1.73 + x / 1.07 - 2x 1.73 + x = .879(1.07-2x) 1.48 + x = .941 - 1.76x -.789 / - 2.76 = .286 = -.286 LOOKING FOR B 1.73 - .286 = 1.44 atm

Consider the following reactions: Eq. 1: A ⇌ B Keq = 1.96x10-4 Eq. 2: B + C ⇌ D Keq = 7.99x103 What is the equilibrium constant for:A + C ⇌ D?

1.57x10^0 Correct. Keq = (1.96x10-4)(7.99x103) = 1.57x10^0

What is the concentration of H3O+ in a 1.62x10^-2 M aqueous solution of HCl?

1.62E-2 Correct. [H3O+] = [HCl] = 1.62x10^-2 M .

Consider a system at 25oC where the following reaction is occurring with Kp = 12.6: 2 A(g) ⇌ B(g) + C(g) At a given instant, the partial pressures at 25oC are the following: Partial pressure of A, atm1.07 Partial pressure of B, atm1.48 Partial pressure of C, atm1.48 Calculate the partial pressure of B at equilibrium.

1.77 atm Correct. Kp=12.6= (1.48 + x)^2 / (1.07 - 2x)^2 SQUARE ROOT 3.55 = 1.48 + x / 1.07 - 2x 1.48 + x = 3.55(1.07-2x) 1.48 + x = 3.80 - 7.1x 2.32 / - 8.1 = -.286 = .286 LOOKING FOR B 1.48 + .286 = 1.77 atm

What is the concentration of H3O+ in a 1.86x10^-3 M aqueous solution of HCl?

1.86E-3 Correct. [H3O+] = [HCl] = 1.86x10^-3 M .

What is the pH of a 1.1x10^-2 M aqueous solution of HNO3?

1.96 Correct. -log(1.1x10^-2) = 1.96

What is the pH of a 1.3x10-4 M aqueous solution of Ba(OH)2?

10.41 Correct. 1.3E-4 * 2 = 2.6x10^-4 14.00 + log(2.6x10^-4) = 10.41

What is the pH of a 4.7x10-4 M aqueous solution of KOH?

10.67 Correct. 4.7x10-4 M KOH * 1 mol OH- / 1 mol KOH = 4.7x10^-4 14.00 + log(4.7x10^-4) = 10.67

What is the pH of a 0.0167 M aqueous solution of NaOH?

12.223 Correct. -log(0.0167) = 1.777 14.0 - 1.777 = 12.223

What is the pH of a 0.0126 M aqueous solution of Ba(OH)2?

12.402 Correct. [OH-] = 2(0.0126) = 0.0252 -log(0.0252) = 1.598 14.000 - 1.598 = 12.402

What is the pH of a 0.0817 M aqueous solution of NaOH?

12.912 Correct. -log(0.0817) = 1.088 14.000 - 1.088 = 12.912

What is the pH of a 2.0x10-1 M aqueous solution of KOH?

13.30 Correct. 2.0x10-1 M KOH * 1 mol OH- / 1 mol KOH = 2.0x10^-1 14.00 + log(2.0x10^-1) = 13.30

Consider the following representation of a equilibrium system where the reaction 2 A(g) ⇌ A2(g)is involved. Assuming each red circle represents 0.00449 atm of A and each pair of blue circle represents 0.00425 atm of A2, what is the value of Kp?

159 Correct. From the figure, one can see that the number of A and A2 circles remains constant after t = 4 minutes indicating that equilibrium was reached at that time. After equilibrium was established, there are 2 red circles representing A and 3 pairs of blue circles representing A2. This means that the partial pressures of A and A2 are PA = 2 * 0.00449 = 0.00898 atm PA2 = 3 * 0.00425 = 0.0128 atm Kp = (PA2) / (PA)^2 Kp = (0.0128) / (0.00898)^2 = 159

Consider a system where the following reaction is occurring: 2 A(s) + 3 B(g) ⇌ 3 C(l) + 2 D(g) The equilibrium partial pressures are: PB, atm0.125 PD, atm0.616 Calculate the value of the equilibrium constant, Kp.

194 Correct. Kp = (PD)^2 / (PB)^3 Kp = (0.616)^2 / (0.125)^3 = 194

Consider a system at 25oC where the following reaction is occurring with Kp = 2.90: 2 A(g) ⇌ B(g) + C(g) At a given instant, the partial pressures at 25oC are: Partial pressure of A, atm0.278 Partial pressure of B, atm0.250 Partial pressure of C, atm0.442 Is the the mixture at equilibrium? If not, which way will the system shift to establish equilibrium? Respond by typing 1, 2, or 3 where 1- the mixture is at equilibrium already 2 - the mixture is not at equilibrium; it will shift right until equilibrium is established 3 - the mixture is not at equilibrium; it will shift left until equilibrium is established

2 Correct. 2 A(g) ⇌ B(g) + C(g) Q = (PB) * (PC) / (PA)^2 Q = (0.250) * (0.442) / (0.278)^2 = 1.43 Comparing Q and K Q = 1.43 < K = 2.90 The answer is 2. The system is not at equilibrium and the reaction will proceed from left to right to increase the partial pressures of the products and decrease those of the reactants until equilibrium is established.

Consider a system at 25oC where the following reaction is occurring with Kp = 1.67: 2 A(g) ⇌ B(g) + C(g) At a given instant, the partial pressures at 25oC are: Partial pressure of A, atm0.352 Partial pressure of B, atm0.104 Partial pressure of C, atm0.476 Is the the mixture at equilibrium? If not, which way will the system shift to establish equilibrium? Respond by typing 1, 2, or 3 where 1- the mixture is at equilibrium already 2 - the mixture is not at equilibrium; it will shift right until equilibrium is established 3 - the mixture is not at equilibrium; it will shift left until equilibrium is established

2 Correct. 2 A(g) ⇌ B(g) + C(g) Q = (PB) * (PC) / (PA)^2 Q= (0.104) * (0.476) / (0.352)^2 Comparing Q and K Q = 0.400 < K = 1.67 The answer is 2. The system is not at equilibrium and the reaction will proceed from left to right to increase the partial pressures of the products and decrease those of the reactants until equilibrium is established.

What is the concentration of OH- in a 2.00x10^-3 M aqueous solution of NaOH?

2.00E-3 Correct. 2.00x10-3 M NaOH * 1 mol OH- / 1 mol NaOH = 2.00x10^-3

At a certain temperature, the equilibrium constant for the chemical equation A(g) + B(g) ⇌ C(g) + 3D(g) is 4.50x10-3. At the same temperature, calculate the equilibrium constant for C(g) + 3D(g) ⇌ A(g) + B(g)

2.22x10^2 Correct. K = 1/Kn = 1/(4.50x10^-3) = 2.22x10^2

What is the pH of a solution where [H3O+] = 2.9x10^-3?

2.54 Correct. -log [2.9x10^-3] = 2.54

What is the pH of a solution where [OH-] = 3.5x10^-12?

2.55 Correct. pH = 14.00 + log[OH-] 14.00 + log[3.5x10-12] = 2.55

Consider a system where the following reaction occurs at constant volume and temperature: 2 A(g) ⇌ B(g) + 2 C(g) The initial partial pressures are: Partial pressure of A, atm 2.02 Partial pressure of B, atm 4.61 Partial pressure of C, atm 8.54 If the partial pressure of B at equilibrium is 2.96 atm, calculate the value of the equilibrium constant, Kp.

2.87 Correct. FIRST: 2.96-4.61 = -1.65 SECOND: 1.65 * 2 = 3.30 THIRD: 2.02 atm 4.61 atm 8.54 atm +3.30 atm - 1.65 atm - 3.30 atm LAST: Kp = (B) (C)^2 / (A)^2 Kp = 2.96 * 5.24^2 / 5.32^2 = 2.87

Consider a system at 25oC where the following reaction is occurring with Kp = 0.0509: 2 A(g) ⇌ 3 B(g) + 2 C(g) At a given instant, the partial pressures at 25oC are: Partial pressure of A, atm0.278 Partial pressure of B, atm0.469 Partial pressure of C, atm0.442 Is the the mixture at equilibrium? If not, which way will the system shift to establish equilibrium? Respond by typing 1, 2, or 3 where 1- the mixture is at equilibrium already 2 - the mixture is not at equilibrium; it will shift right until equilibrium is established 3 - the mixture is not at equilibrium; it will shift left until equilibrium is established

3 Correct. 2 A(g) ⇌ 3 B(g) + 2 C(g) Q= (PB)^3 * (PC)^2 / (PA)^2 (0.469)^3 *(0.442)^2 / (0.278)^2 = 0.261 Comparing Q and K (Q = 0.261) > (K = 0.0509) The answer is 3. The system is not at equilibrium and the reaction will proceed from right to left to decrease the partial pressures of the products and increase those of the reactants until equilibrium is established.

Consider a system at 25oC where the following reaction is occurring with Kp = 0.0646: 2 A(g) ⇌ 3 B(g) + 2 C(g) At a given instant, the partial pressures at 25oC are: Partial pressure of A, atm0.327 Partial pressure of B, atm0.574 Partial pressure of C, atm0.365 Is the the mixture at equilibrium? If not, which way will the system shift to establish equilibrium? Respond by typing 1, 2, or 3 where 1- the mixture is at equilibrium already 2 - the mixture is not at equilibrium; it will shift right until equilibrium is established 3 - the mixture is not at equilibrium; it will shift left until equilibrium is established

3 Correct. 2 A(g) ⇌ 3 B(g) + 2 C(g) Q= (PB)^3 * (PC)^2 / (PA)^2 Q= (0.574)^3 * (0.365)^2 / (0.327)^2 = 0.236 Comparing Q and K (Q = 0.236) > (K = 0.0646) The answer is 3. The system is not at equilibrium and the reaction will proceed from right to left to decrease the partial pressures of the products and increase those of the reactants until equilibrium is established.

Consider the following reactions: Eq. 1: A ⇌ B Keq = 1.04x10-4 Eq. 2: B + C ⇌ D Keq = 2.92x105 What is the equilibrium constant for: A + C ⇌ D?

3.04x10^1 Correct. Keq = (1.04x10^-4)(2.92x10^5) = 3.04x10^1

At a certain temperature, the equilibrium constant for the chemical equation A(g) + B(g) ⇌ C(g) + D(g) is 6.76x10-3. At the same temperature, calculate the equilibrium constant for 3A(g) + 3B(g) ⇌ 3C(g) + 3D(g)

3.09x10^-7 Correct. Coefficient factor, n, of 3. By using the coefficient rule, we can calculate the K for this equation. K' = Kn= (6.76x10^-3)^3 = 3.09x10^-7

At a certain temperature, the equilibrium constant for the chemical equation HC2H3O2(aq) + H2O(l) ⇌ H3O+(aq) + C2H3O2-(aq) is 1.80x10-5. At the same temperature, calculate the equilibrium constant for 2HC2H3O2(aq) + 2H2O(l) ⇌ 2H3O+(aq) + 2C2H3O2-(aq)

3.24E-10 Correct. K = Kn= (1.80x10^-5)^2 = 3.24x10^-10

At a certain temperature, the equilibrium constant for the chemical equation HC2H3O2(aq) + H2O(l) ⇌ H3O+(aq) + C2H3O2-(aq) is 1.80x10^-5. At the same temperature, calculate the equilibrium constant for 2HC2H3O2(aq) + 2H2O(l) ⇌ 2H3O+(aq) + 2C2H3O2-(aq)

3.24x10^-10 Correct. K = Kn = (1.80x10^-5)^2 = 3.24x10^-10 WE SQUARED IT BECAUSE OF COEFFICIENT FROM BOTTOM FORMULA

Consider the hypothetical reaction: A(s) ⇌ B(aq) + 2 C(aq) If at a given instant, the concentrations of B and C are 2.53x10-8 and 3.64x10-8 mol/L, the value of the reaction quotient (Q) is

3.35x10^-23 Correct. Q = [B][C]^2 / 1 = (2.53x10^-8)(3.64x10-8)^2 = 3.35x10^-23

Which of these would be the pH of a 1.69x10^-4 M solution of HX(aq) if HX is a strong acid?

3.771 Correct. -log(1.69x10^-4) = 3.771

What is the pH of a 1.3x10^-4 M aqueous solution of HBr?

3.88 Correct. -log(1.3x10^-4) = 3.88

Consider the reaction: N2 + 3 H2 ⇌ 2 NH3 If at a given instant, the partial pressures of N2, H2, and NH3 are 10.6, 37.7, and 19.7 Torr. Calculate the reaction quotient (Q).

3.95x10^2 Correct. Partial pressure of N2 = 10.6 Torr or 1.395x10^-2 atm. Therefore, activity of N2 = 1.395x10^-2. Partial pressure of H2 = 37.7 Torr or 4.960x10-2 atm. Therefore, activity of H2 = 4.960x10-2. Partial pressure of NH3 = 19.7 Torr or 2.592x10-2 atm. Therefore, activity of NH3 = 2.592x10-2. Finally, Q = (2.592x10-2)^2 / [ (1.395x10-2)(4.960x10-2)^3 ] = 3.95x10^2

At a certain temperature, the equilibrium constant for the chemical equation 3A(g) + B(g) ⇌ 3C(g) + 2 D(g)is 5.68x10^-2. At the same temperature, calculate the equilibrium constant for 6C(g) + 4D(g) ⇌ 6A(g) + 2B(g)

310 Correct. 1/(5.68x10^-2)^2 = 310

Which of these would be the pH of a 9.54x10^-5 M solution of HX(aq) if HX is a strong acid?

4.020 Correct. -log(9.54x10^-5) = 4.020

At a certain temperature, the equilibrium constants are known for the chemical equations Ag2CO3(s) ⇌ 2 Ag+(aq) + CO32-(aq) K = 8.00x10^-12 Ag+(aq) + 2 NH3(aq) ⇌ Ag(NH3)2+(aq) K = 1.70x10^7 At the same temperature, calculate the equilibrium constant for CO32-(aq) + 2 Ag(NH3)2+(aq) ⇌ 4 NH3(aq) + Ag2CO3(s)

4.32x10^-4 (PUT IN .000432) Correct. 2 Ag+(aq) + CO32-(aq) ⇌ Ag2CO3(s) K' = 1 / 8.00x10-12 = 1.25x10^11 2 Ag(NH3)2+(aq) ⇌ 2 Ag+(aq) + 4 NH3(aq) K' = 1 / (1.70x107)^2 = 3.46x10^-15 K = 1.25x10^11 * 3.46x10^-15 = 4.32x10^-4

What is the [OH-] in a solution with a pH of 4.64?

4.4E-10 Correct. 14.00 - 4.64 = 9.36 [OH-] = 10-pOH = 10^-9.36 = 4.4x10^-10 M

Consider the following reactions: Eq. 1: A ⇌ B Keq = 6.41x10-3 Eq. 2: B + C ⇌ D Keq = 1.15x104 What is the equilibrium constant for:2A + C ⇌ B + D?

4.72x10^-1 Correct. Keq = (6.41x10-3)^2(1.15x104) = 4.72x10^-1

Suppose the reaction quotient (Q) for a reaction mixture at a particular instant is 0.675 if the reaction is written as 2 NH3 → N2 + 3 H2 Calculate the reaction quotient if the reaction is written as 4 N2 + 12 H2 → 8 NH3.

4.82 Correct. Q = Reciprocal of (original Q)^4 = 1/(0.675)^4 = 4.82

Consider the reaction: N2 + 3 H2 ⇌ 2 NH3 If at a given instant, the partial pressures of N2, H2, and NH3 are 17.7, 43.4, and 36.1 Torr. Calculate the reaction quotient (Q).

5.20x10^2 Correct. Partial pressure of N2 = 17.7 Torr or 2.329x10-2 atm. Therefore, activity of N2 = 2.329x10-2. Partial pressure of H2 = 43.4 Torr or 5.710x10-2 atm. Therefore, activity of H2 = 5.710x10-2. Partial pressure of NH3 = 36.1 Torr or 4.750x10-2 atm. Therefore, activity of NH3 = 4.750x10-2. Finally, Q = (4.750x10-2)2 / [ (2.329x10-2)(5.710x10-2)3 ] = 5.20x102

Suppose the reaction quotient (Q) for a reaction mixture at a particular instant is 2.30 if the reaction is written as N2 + 3 H2 → 2 NH3 Calculate the reaction quotient if the reaction is written as 2 N2 + 6 H2 → 4 NH3.

5.29 Correct. Q = (original Q)^2 = (2.30)^2 = 5.29 BOTTOM QUOTIENT IS 2 TIMES MORE THAN TOP

Consider the reaction: N2 + 3 H2 = 2 NH3If at a given instant, the partial pressures of N2, H2, and NH3 are 14.0, 28.3, and 17.5 Torr. Calculate the reaction quotient (Q).

5.58x10^2 Correct. Partial pressure of N2 = 14.0 Torr or 1.842x10^-2 atm. Therefore, activity of N2 = 1.842x10-2. Partial pressure of H2 = 28.3 Torr or 3.724x10^-2 atm. Therefore, activity of H2 = 3.724x10-2. Partial pressure of NH3 = 17.5 Torr or 2.303x10^-2 atm. Therefore, activity of NH3 = 2.303x10-2. Q = (2.303x10-2)^2 / [ (1.842x10-2)(3.724x10-2)^3 ] = 5.58x10^2

What is the pOH of a solution where [OH-] = 1.0x10^-6?

6.00 Correct. -log10[1.0x10-6] = 6.00

Consider the following reactions: 3 A(aq) ⇌ 2 B(aq) Keq = x 6 A(aq) ⇌ 4 B(aq) Keq = y What is the equilibrium constant for the second reaction if the equilibrium for the first reaction (x) is 2.45x10^-3?

6.00x10^-6 Correct. If we double the coefficients a chemical equation, the equilibrium constant for the resulting reaction is just the square of the original equilibrium constant: y = x^2 = (2.45x10-3)^2 = 6.00x10^-6

Consider the following reactions: A(aq) ⇌ 2 B(aq) Keq = x 2 A(aq) ⇌ 4 B(aq) Keq = y What is the equilibrium constant for the second reaction if the equilibrium for the first reaction (x) is 2.95x10^-5?

8.70x10^-10 Correct. If we double the coefficients a chemical equation, the equilibrium constant for the resulting reaction is just the square of the original equilibrium constant: y = x^2 = (2.95x10-5)^2 = 8.70x10^-10

What is the concentration of OH- in a 4.46x10-4 M aqueous solution of Ca(OH)2?

8.92E-4 Correct. 4.46E-4 * 2 mol OH- / 1 mol Ca(OH)2 = 8.92x10^-4 M OH-

What is the [H3O+] in a solution where [OH-] is 1.10x10-8?

9.12E-7 Correct. [H3O+]= 1.00 x 10^-14 / 1.10x10^-8 = 9.12x10^-7 M

Consider a system where the following reaction occurs at constant volume and temperature: 2 A(g) ⇌ B(g) + C(g) The initial partial pressures are: Partial pressure of A, atm 0.830 Partial pressure of B, atm 0.696 Partial pressure of C, atm 0.409 If the partial pressure of A at equilibrium is 0.271 atm, calculate the value of the equilibrium constant, Kp.

9.16 Correct. FIRST: .271-.830 = -.559 SECOND: .559/2 = .280 THIRD: 0.830 atm 0.696 atm 0.409 atm -0.559 atm + 0.280 atm + 0.280 atm LAST: Kp = (B) (C) / (A)^2 Kp = .976 * .689 / .271^2 = .916

Consider the following reactions: 3X(aq) + Y(aq) ⇌ 2Z(aq) Keq = x 2Z(aq) ⇌ 3X(aq) + Y(aq) Keq = y What is the equilibrium constant for the second reaction if the equilibrium for the first reaction (x) is 1.04x10^-4?

9.62x10^3 Correct. If we flip a chemical equation, the equilibrium constant for the resulting reaction is just the reciprocal of the original equilibrium constant: y = 1/x = 1/1.04x10-4 = 9.62x10^3 REACTION ARE FLIPPED THAT IS WHY WE DIVIDE!

Which of the following is the conjugate base of HCO3-?

CO32- Correct. The Bronsted-Lowry model considers that an acid is a proton donor and a base is a proton acceptor.The species formed when a proton is removed from an acid is referred to as the conjugate base of that acid, i.e. B- is the conjugate base of HB. The species formed when a proton is added to a base is called the conjugate acid of that base, i.e. HA is the conjugate acid of A-.

True or False. At equilibrium, all atomic and molecular motions cease to occur in the system.

False

Which of the following is the conjugate acid of HCO3-?

H2CO3

Consider the following reaction: CN-(aq) + H2O(l) ⇌ HCN(aq) + OH-(aq) which occurs when potassium cyanide (KCN) is dissolved in water. The Bronsted-Lowry acid for the forward reaction is

H2O

Which of the following is the conjugate acid of HSO4-?

H2SO4

Which of the following is the conjugate acid of CO32-?

HCO3-

Which of the following is the conjugate base of H2CO3?

HCO3-

Which of the following is the conjugate acid of SO42-?

HSO4-

Which of the following is the conjugate base of H2SO4?

HSO4-

Consider the following representation of a fictitious reaction 2 A ⇌ A2 Which of the following is/are true? I. The reaction is not equilibrium at t=2 minutes but is at equilibrium at t=4 minutes. II. The reaction is at equilibrium at t>2 minutes. III. The reaction is not at equilibrium at t=0 minutes. IV. The forward and reverse reaction rates are equal at t=6 minutes.

I, III, and IV only Correct. For t=4 minutes and greater, we see that there are 2 A and 3 A2 molecules. Since the concentration of reactants and products are not changing, the system is at equilibrium for t=4 minutes and above. Chemical equilibrium is dynamic. The reason we do not observe changes at equilibrium is NOT because nothing's happening, but because both forward and reverse reactions are occurring at the same rate.

Which of the following is/are true about a reaction mixture at equilibrium? I. The concentration of products are equal to the concentration of reactants. II. Both forward and reverse reactions have stopped. III. The forward and reverse reactions are occurring at equal rates. IV. The concentration of reactants and products are not changing.

III and IV only

Consider the following reaction: NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq) which occurs when ammonium chloride (NH4Cl) is dissolved in water. The Bronsted-Lowry base for the reverse reaction is

NH3

A student is studying the equilibrium represented by the equation: [Cu(H2O)6]2+(aq, blue) + 5Br-(aq) ⇌ CuBr42-(aq, green) + 6H2O(l) The equilibrium mixture is blue in color. The student draws some of the blue solution into the bulb of a Beral pipet. The pipet is placed, bulb down, in a hot-water bath. After a short time, the solution changes from blue to green. From these observations can we conclude that heating favors the reverse reaction vs. the forward reaction?

Mg(OH)2 is insoluble in water, which means that the equilibrium constant for Mg(OH)2(s) ⇌ Mg2+(aq) + 2 OH-(aq) is a very small number. To dissolve Mg(OH)2, we need to shift the equilibrium to the right. Can this be done by adding NaOH(aq)?

NO2(g) has a reddish color. N2O4(g) is colorless. In an equilibrium mixture of these two gases, the following reaction occurs: 2 NO2(g) ⇌ N2O4(g)When a reddish-colored sample of this mixture is cooled, the color fades. From this observation, can we conclude that Keq for the reaction increases as the temperature increases?

No Correct.

NO2(g) has a reddish color. N2O4(g) is colorless. In an equilibrium mixture of these two gases, the following reaction occurs: 2 NO2(g) ⇌ N2O4(g)When a reddish-colored sample of this mixture is cooled, the color fades. From this observation, can we conclude that heating favors the forward reaction vs. the reverse the reaction?

No Correct. Heating favors the endothermic direction. The given information tells us that the forward direction (less red) is favored by cooling, not heating. Therefore the forward direction is exothermic, the reverse is endothermic. Less red at equilibrium means lower Keq at equilibrium.

No Correct. Heating favors the endothermic direction. The given information tells us that the forward direction (more green color vs. blue) is favored by heating. Therefore the forward direction is endothermic, the reverse is exothermic. More green at equilibrium means higher Keq at equilibrium.

Which of the following is the conjugate base of HSO4-?

SO4^2-

State whether HBr is a strong or weak acid.

Strong Correct.

State whether HNO3 is a strong or weak acid.

Strong Correct.

State whether H2SO4 is a strong or weak acid.

Strong Correct. H2SO4 is a strong acid.You must learn the strong acids to be able to determine what is strong and what is weak. The strong acids are HCl, HBr, HI, HClO4, H2SO4, and HNO3.

State whether LiOH is a strong or weak base.

Strong Correct. LiOH is a strong base. You must learn the strong bases to be able to determine what is strong and what is weak. The strong bases are LiOH, NaOH, KOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2.

State whether HI is a strong or weak acid.

Strong. Correct. HI is a strong acid.You must learn the strong acids to be able to determine what is strong and what is weak. The strong acids are HCl, HBr, HI, HClO4, H2SO4, and HNO3.

True or False. At equilibrium, the rate of formation of products is equal to the rate of formation of reactants.

True

True or False. Once a reaction reaches equilibrium, the concentrations of reactants and products remain constant.

True

True or False. When a reaction reaches equilibrium, the forward and reverse reactions are occurring at the same rate.

True

A student is studying the equilibrium represented by the equation:[Cu(H2O)6]2+(aq, blue) + 5Br-(aq) ⇌ CuBr42-(aq, green) + 6H2O(l)The equilibrium mixture is blue in color.The student draws some of the blue solution into the bulb of a Beral pipet. The pipet is placed, bulb down, in a hot-water bath. After a short time, the solution changes from blue to green. From these observations can we conclude that heating favors the forward reaction vs. the reverse the reaction?

Yes

NO2(g) has a reddish color. N2O4(g) is colorless. In an equilibrium mixture of these two gases, the following reaction occurs: 2 NO2(g) ⇌ N2O4(g) When a reddish-colored sample of this mixture is cooled, the color fades. From this observation, can we conclude that Keq for the reaction decreases as the temperature increases?

Yes

NO2(g) has a reddish color. N2O4(g) is colorless. In an equilibrium mixture of these two gases, the following reaction occurs: 2 NO2(g) ⇌ N2O4(g) When a reddish-colored sample of this mixture is cooled, the color fades. From this observation, can we conclude that the forward reaction is exothermic?

Yes

A student is studying the equilibrium represented by the equation: [Cu(H2O)6]2+(aq, blue) + 5Br-(aq) ⇌ CuBr42-(aq, green) + 6H2O(l) The equilibrium mixture is blue in color. The student draws some of the blue solution into the bulb of a Beral pipet. The pipet is placed, bulb down, in a hot-water bath. After a short time, the solution changes from blue to green. From these observations can we conclude that Keq for the reaction increases as the temperature increases?

Yes Correct. Heating favors the endothermic direction. The given information tells us that the forward direction (more green color vs. blue) is favored by heating. Therefore the forward direction is endothermic, the reverse is exothermic. More green at equilibrium means higher Keq at equilibrium.

For the following pair of acids: acid 1: H2CO3 acid 2: HCO3- which is the stronger acid?

acid 1

For the following pair of acids: acid 1: H2S acid 2: H2O which is the stronger acid?

acid 1

For the following pair of acids: acid 1: HF acid 2: H2O which is the stronger acid?

acid 1

For the following pair of acids: acid 1: HI acid 2: HBr which is the stronger acid?

acid 1

For the following pair of acids: acid 1: H3PO4 acid 2: H3PO3 which is the stronger acid?

acid 1 Correct. For a series of oxoacids, the acid strength increases with the number of O atoms bonded to Y.

For the following pair of acids: acid 1: HClO acid 2: HClO2 which is the stronger acid?

acid 2

For the following pair of acids: acid 1: H2S acid 2: H2Te which is the stronger acid?

acid 2 Correct. In going down a column of elements, the size of atom X increase, the H-X bond strength decreases, and the strength of the binary acid increases.

For the following pair of acids: acid 1: HSO4- acid 2: H2SO4 which is the stronger acid?

acid 2 Correct. acid strength decreases with negative anion charge

For the following pair of acids: acid 1: HIO acid 2: HClO which is the stronger acid?

acid 2 Correct. oxoacids of the same structure (H-O-Y-), differing only in the atom Y, has acid strength increasing with electronegativity of Y

For the following pair of acids: acid 1: H2Te acid 2: HI which is the stronger acid?

acid 2. Correct. Going across a row of elements, the electronegativity increase, the H-X bond polarity increases, and the acid strength increases.

Consider the following fictitious reaction at equilibrium: A(s) + B(g) ⇌ C(g) ΔH = -80 kJ Which of the following stresses would not shift the equilibrium?

adding more A

Consider the following fictitious reaction at equilibrium: A(g) + B(g) ⇌ C(g) ΔH = -80 kJ Which of the following stresses would shift the equilibrium to the left?

adding more C

Which is the weaker base? Base 1. ClO3- Base 2. ClO2-

base 1

Which is the weaker base? Base 1. ClO4- Base 2. ClO3-

base 1

Which is the weaker base? Base 1. BrO3- Base 2. BrO-

base 1 Correct. For oxoanions with the same central atom, the more oxygens attached to the central atom, the more the charge is dispersed, the more stable the anion, the lesser is its tendency to accept to a proton, the weaker the base.

Which of the following describes the effect of changing the temperature of a reaction mixture at equilibrium? (Q is the reaction quotient, Keq is the equilibrium constant)

change in Keq; reaction shifts to the left or right until Q=Keq again

Which of the following describes the effect of cooling a reaction mixture at equilibrium, if the reverse reaction is exothermic? (Keq is the equilibrium constant)

decrease in Keq; reaction shifts to the left

Which of the following describes the effect of heating a reaction mixture at equilibrium, if the forward reaction is exothermic? (Keq is the equilibrium constant)

decrease in Keq; reaction shifts to the left

Which of the following describes the effect of heating a reaction mixture at equilibrium, if the reverse reaction is endothermic? (Keq is the equilibrium constant)

decrease in Keq; reaction shifts to the left Correct. A temperature increase favors the endothermic direction. If the forward direction is endothermic, then a higher temperature means a higher Keq and a lower temperature means a lower Keq. Similarly, if the forward reaction is exothermic, then a higher temperature means a lower Keq and a lower temperature means a higher Keq.

Which of the following describes the effect of removing products from a reaction mixture at equilibrium? (Q is the reaction quotient, Keq is the equilibrium constant)

decrease in Q; reaction shifts to the right until Q=Keq again

The oxidation of carbon monoxide proceeds as follows: 2CO (g) + O2(g) ⇌ 2CO2(g) ΔH = -559 kJ Which of the following will cause the equilibrium to shift to the left?

decreasing the pressure of the system at constant temperature

Consider the following fictitious reaction at equilibrium:A(g) + B(g) ⇌ C(g) ΔH = -80 kJ Which of the following stresses would shift the equilibrium to the right?

decreasing the temperature

Which of the following describes the effect of heating a reaction mixture at equilibrium, if the forward reaction is endothermic? (Keq is the equilibrium constant)

increase in Keq; reaction shifts to the right

Which of the following describes the effect of heating a reaction mixture at equilibrium, if the reverse reaction is exothermic? (Keq is the equilibrium constant)

increase in Keq; reaction shifts to the right

Which of the following describes the effect of cooling a reaction mixture at equilibrium, if the forward reaction is exothermic? (Keq is the equilibrium constant)

increase in Keq; reaction shifts to the right Correct.

Which of the following describes the effect of cooling a reaction mixture at equilibrium, if the reverse reaction is endothermic? (Keq is the equilibrium constant)

increase in Keq; reaction shifts to the right Correct. A temperature increase favors the endothermic direction. If the forward direction is endothermic, then a higher temperature means a higher Keq and a lower temperature means a lower Keq. Similarly, if the forward reaction is exothermic, then a higher temperature means a lower Keq and a lower temperature means a higher Keq.

Which of the following describes the effect of adding products to a reaction mixture at equilibrium? (Q is the reaction quotient, Keq is the equilibrium constant)

increase in Q; reaction shifts to the left until Q=Keq again

Which of the following describes the effect of removing reactants from a reaction mixture at equilibrium? (Q is the reaction quotient, Keq is the equilibrium constant)

increase in Q; reaction shifts to the left until Q=Keq again

Consider the following fictitious reaction at equilibrium: A(g) + B(g) ⇌ C(g) ΔH = +25 kJ Which of the following stresses would shift the equilibrium to the right?

increasing the temperature

Consider the following fictitious reaction at equilibrium:A(g) + B(g) ⇌ C(g) ΔH = -80 kJ Which of the following stresses would shift the equilibrium to the left?

increasing the temperature

Consider the following fictitious reaction at equilibrium: A(s) + B(g) ⇌ C(g) ΔH = -80 kJ Which of the following stresses would not shift the equilibrium?

increasing the volume by decreasing the pressure at constant temperature

A student is studying the equilibrium represented by the equation: 2CrO42-(aq, yellow) + 2 H3O+(aq) ⇌ Cr2O72-(aq, orange) + 3H2O(l) The equilibrium mixture is yellow in color. What direction will the equilibrium shift and what color should the mixture be after adding NaOH(s) to the equilibrium mixture?

left; more yellow

A student is studying the equilibrium represented by the equation: [Cu(H2O)6]2+(aq, blue) + 5Br-(aq) ⇌ CuBr42-(aq, green) + 6H2O(l) The equilibrium mixture is blue in color. What direction will the equilibrium shift and what color should the mixture be after adding NaBr(s) to the equilibrium mixture?

right; more green

right; more orange

right; more yellow

Consider a hypothetical acid-base indicator, "HX", ionizes in water according to the equation HX(aq) ⇌ H+(aq) + X-(aq) Suppose HX(aq) is red and X- is yellow. How would an equilibrium mixture behave if we were to add concentrated HCl?

shift left; more red

shift right; more yellow

A reaction has reached equilibrium and the concentration of reactant "X" was found to be 2.22x10-5 mol/L. If the system is left undisturbed, we expect

the concentration of X to remain at 2.22x10-5 mol/L

A reaction has reached equilibrium and the concentration of reactant "X" was found to be 2.69x10-5 mol/L. If the system is left undisturbed, we expect

the concentration of X to remain at 2.69x10-5 mol/L

A reaction has reached equilibrium and the concentration of reactant "X" was found to be 4.56x10^-5 mol/L. If the system is left undisturbed, we expect

the concentration of X to remain at 4.56x10^-5 mol/L

A reaction has reached equilibrium and the concentration of reactant "X" was found to be 4.87x10-5 mol/L. If the system is left undisturbed, we expect

the concentration of X to remain at 4.87x10-5 mol/L

Consider the hypothetical reaction: A(s) ⇌ B(aq) + 2 C(aq) If at a given instant, the concentrations of B and C are 2.28x10-8 and 0.786 mol/L, and the equilibrium constant (Keq) is 4.74x10-23, then

the reaction quotient Q is larger than Keq and we expect formation of more A(s) until Q=Keq Correct. Q is larger than Keq. There will be a net reverse reaction (formation of A) until equilibrium is established (Q=Keq)

Consider the hypothetical reaction: A(s) ⇌ B(aq) + 2 C(aq) If at a given instant, the concentrations of B and C are 2.78x10-8 and 0.388 mol/L, and the equilibrium constant (Keq) is 8.59x10^-23, then

Consider the hypothetical reaction: A(s) ⇌ B(aq) + 2 C(aq) If at a given instant, the concentrations of B and C are 3.02x10-8 and 0.589 mol/L, and the equilibrium constant (Keq) is 1.10x10-22, then

CHEM - 2

Related study sets

NSG 356 EXAM 3!

Ch 66 PrepU: Management of Patients With Neurologic Dysfunction

UNIT 1

2023 TITAN XD CERTIFICATION

Acid Base Balance - WSU N320

Words2

Banking

Blood cancers

Craven Ch 19: Vital Signs

Exam 2 Psychology

Practice Questions

biology exam FINAL

Ch. 10, 11, 12, 13

GCSE History American West

Lauren Sousa History midterm Hanna Review

Abnormal Psych Chapter 10

MODULE 4

Chapter 2

Kapalan LTC Review

Promulgated Contract Forms