MCAT General Chemistry Discrete Question Set 20: Kinetics & Equilibrium

Transcription

1 MCAT General Chemistry Discrete Question Set 0: Kinetics & Equilibrium Question No. 1 of 10 Instruction: (1) Read the problem statement and answer choices carefully () Work the problems on paper as needed (3) 1: If the rate law for a reaction is Rate = k[no ], which of the elementary steps in the proposed mechanism is the slowest? Step 1: NO + NO NO 3 + NO Question #01 Step : NO 3 + CO NO + CO (A) Step 1 (B) Step (C) They re the same. (D) Can t determine A: Correct. The slowest step in the reaction mechanism. The rate of the slowest step determines the rate of the overall reaction. B: Incorrect. The slowest step in the reaction mechanism. The rate of the slowest step determines the rate of the overall reaction. C: Incorrect. The slowest step in the reaction mechanism. The rate of the slowest step determines the rate of the overall reaction. D: Incorrect. The slowest step in the reaction mechanism. The rate of the slowest step determines the rate of the overall reaction. The slowest step in the reaction mechanism. The rate of the slowest step determines the rate of the overall reaction The rate law has NO to the nd order. Therefore, NO molecules must collide in the rate determining step. The rate determining step is the one that has NO molecules as reactions Step 1 does this Answer: (A) Step 1

2 Question No. of 10 Instruction: (1) Read the problem statement and answer choices carefully () Work the problems on paper as needed (3) : Based on the following data, write the complete rate law? Question #0 [NO] (in M) [O ] (in M) Initial rate (in M/sec) (A) Rate = k[no] [O ] (B) Rate = k[no][o ] Rate = k[no][o ] (D) Rate = k[no] [O ] A: Correct. As [NO] is doubled, the rate goes up by 4 and as [O] is doubled, the rate is doubled. B: Incorrect. As [NO] is doubled, the rate goes up by 4 and as [O] is doubled, the rate is doubled. C: Incorrect. As [NO] is doubled, the rate goes up by 4 and as [O] is doubled, the rate is doubled. D: Incorrect. As [NO] is doubled, the rate goes up by 4 and as [O] is doubled, the rate is doubled. As [NO] is doubled (between trial 1 & ), the rate goes up by 4. NO is nd order As [O ] is doubled (between trial 1 & 4), the rate is doubled. O is 1 st order Rate = k[no] [O ] Answer: (A) Rate = k[no] [O ]

3 Question No. 3 of 10 Instruction: (1) Read the problem statement and answer choices carefully () Work the problems on paper as needed (3) 3: What is the correct equilibrium constant expression for the following reaction? NO Cl (g) NO (g) + Cl (g) A. K = [ NO ][ Cl ] [ NO Cl] Question #03 B. K = [ NOCl] [ NO ][ Cl ] C. D. [ NOCl] [ NO ] [ Cl ] [ NO ] [ Cl ] [ NO Cl] A: Incorrect. The equilibrium constant expression has the concentration of the B: Incorrect. The equilibrium constant expression has the concentration of the C: Incorrect. The equilibrium constant expression has the concentration of the D: Correct. The equilibrium constant expression has the concentration of the [ NO ] [ Cl ] Answer: (D) [ NO Cl]

4 Question No. 4 of 10 Instruction: (1) Read the problem statement and answer choices carefully () Work the problems on paper as needed (3) 4: Which of the changes will shift the reaction to the right when disturbing equilibrium for N (g) + 3 H (g) NH 3 (g) kj? I. Increasing temperature II. Decreasing temperature III. Increasing volume IV. Decreasing volume V. Removing NH 3 VI. Adding NH 3 Question #04 VII. Removing N VIII. Adding N (A) I, IV, VI, VII (B) II, III, V, VIII (C) I, VI, VIII (D) II, IV, V, VIII A: Incorrect. Removing N is removing a reactant that would push the reaction to the left. B: Incorrect. Increasing volume lowers the pressure. The reaction will shift to the side with the greatest number of gas molecules to increase the pressure again. For this reaction, it would shift to the left. C: Incorrect. Increasing temperature for this reaction is an increase in a product that would push the reaction to the left. D: Correct. You chose all the options that would push the reaction to the right! I. Energy is a product. Increasing a product would push the reaction to the left. II. Energy is a product. Decreasing a product would push the reaction to the right III. Increasing volume pushes towards side with more moles of gas. The left has 4, the right has. It would push reaction towards left. IV. Decreasing volume pushes towards side with least moles of gas. It would push the reaction towards the right. V. Removing a product shifts the reaction towards the right VI. Adding a product shifts the reaction to the left VII. Removing a reactant shifts the reaction to the left VIII. Adding a reactant shifts the reaction to the right II, IV, V, VIII Answer: (D) II, IV, V, VIII

5 Question No. 5 of 10 Instruction: (1) Read the problem statement and answer choices carefully () Work the problems on paper as needed (3) 5: According to Collision Theory, what is necessary for a reaction to occur? Question #05 (A) Collision (B) Collision with correct orientation (C) Collision with activation energy (D) Collision with correct orientation and activation energy A: Incorrect. The molecules must collide, but that s not all. B: Incorrect. The molecules must collide with the correct orientation but there is another requirement as well. C: Incorrect. The molecules must collide with the activation energy, but there is another requirement as well. D: Correct. The molecules must collide with the correct orientation and the minimum energy (activation energy). Answer: (D) Collision with correct orientation and activation energy

6 Question No. 6 of 10 Instruction: (1) Read the problem statement and answer choices carefully () Work the problems on paper as needed (3) 6: A higher activation energy leads to a reaction rate. Question #06 (A) higher (B) lower (C) same (D) there is no relationship between the two A: Incorrect. The higher the activation energy, the less likely a collision with that minimum energy will occur. B: Correct. The higher the activation energy, the less likely a collision with that minimum energy will occur. C: Incorrect. The higher the activation energy, the less likely a collision with that minimum energy will occur. D: Incorrect. The higher the activation energy, the less likely a collision with that minimum energy will occur. The higher the activation energy, the lower the chance that a collision will have that minimum energy. The lower the chance of a successful collision, the lower the reaction rate Answer: (B) lower

7 Question No. 7 of 10 Instruction: (1) Read the problem statement and answer choices carefully () Work the problems on paper as needed (3) 7: H + I HI Write the equilibrium constant expression for this all gas reaction? Question #07 (A) (B) (C) (D) [ H ][ I ] [ HI] [ H ] [ I] [ HI] [ HI] [ H ][ I ] [ HI] [ H ] [ I] A: Incorrect. The equilibrium constant expression has the concentration of the B: Incorrect. The equilibrium constant expression has the concentration of the C: Correct. The equilibrium constant expression has the concentration of the D: Incorrect. The equilibrium constant expression has the concentration of the Ratio of products to reactants using balanced equation coefficients as powers [ HI] Answer: (C) [ H ][ I ]

8 Question No. 8 of 10 Instruction: (1) Read the problem statement and answer choices carefully () Work the problems on paper as needed (3) 8: If the equilibrium constant K for the reaction A B + C is 5, what is the equilibrium constant for the reverse reaction at the same temperature? Question #08 (A) 6 (B) 1/5 (C) 1/6 (D) Cannot be determined A: Incorrect. When you flip a reaction, take the inverse of the equilibrium constant. B: Correct. When you flip a reaction, take the inverse of the equilibrium constant. C: Incorrect. When you flip a reaction, take the inverse of the equilibrium constant. D: Incorrect. When you flip a reaction, take the inverse of the equilibrium constant. When reversing the reaction, take the inverse of the equilibrium constant. Answer: B. 1/5

9 Question No. 9 of 10 Instruction: (1) Read the problem statement and answer choices carefully () Work the problems on paper as needed (3) 9: Catalysts increase rate of reaction by: Question #09 (A) Decreasing energy change of reaction (B) Decreasing potential energy of reactants (C) Decreasing potential energy of products (D) Decreasing activation energy A: Incorrect. Catalysts don t change the overall energy change of the reaction. B: Incorrect. Catalysts don t change the reactants themselves. C: Incorrect. Catalysts don t change the products themselves. D: Correct. Catalysts decrease activation energy. Catalysts do not change the reactants or products, therefore options A, B and C are incorrect. Catalysts change the path between the reactants and products the activation energy would change option D is correct. A catalysts cannot increase the number of collisions option E is incorrect. Answer: D. Decreasing activation energy

10 Question No. 10 of 10 Instruction: (1) Read the problem statement and answer choices carefully () Work the problems on paper as needed (3) 10: For NO + O NO, the proposed mechanism would support which rate law: Step 1: NO + NO N O 4 slow Step : N O + O NO fast Question #10 (A) Rate = k[no] (B) Rate = k[no]/[o ] (C) Rate = k[no] /[O ] (D) Rate = k[no] [O ] A: Correct. The rate law can be written from the slowest elementary step in the reaction mechanism. B: Incorrect. The rate law can be written from the slowest elementary step in the reaction mechanism. C: Incorrect. The rate law can be written from the slowest elementary step in the reaction mechanism. D: Incorrect. The rate law can be written from the slowest elementary step in the reaction mechanism. The rate law cannot be written from the overall balanced equation, but it can be written from the slowest step of the reaction mechanism. NO + NO N O 4 slow This step would give a rate law that is second order in NO Answer: A. Rate = k[no]