Expert Chemistry Tutoring

MS1.

- State of equilibrium where the forward and reverse reactions occur at the same rate. (1 mark)

MS2.

- Catalyst speeds up both forward and reverse reactions equally. (1 mark)

- Equilibrium is reached more quickly. (1 mark)

MS3.

- Principle: If a change is made to the conditions of a system at equilibrium, the system will shift in a direction that reduces the effect of the change. (1 mark)

- Example: Adding heat to an exothermic reaction will shift the equilibrium to the left (i.e., favor the endothermic reverse reaction). (1 mark)

MS4.

- Shifts equilibrium to the right (favors NO2 production). (1 mark)

- Le Chatelier's principle: NO is a reactant, so increasing its concentration shifts the equilibrium towards the products. (1 mark)

- According to the stoichiometry, 2 moles of NO react with 1 mole of O2 to produce 2 moles of NO2. (1 mark)

MS5.

- Correct equation: N2 + 3H2 ⇌ 2NH3 (4 marks)

MS6.

- Kc = [HI]^2 / [H2] [I2] (1 mark)

- Kc = (0.75)^2 / (0.20) (0.05) = 22.5 (2 marks)

MS7.

- Contains a weak acid and its conjugate base or a weak base and its conjugate acid. (1 mark)

- Neutralizes added acid/base without significantly changing pH. (1 mark)

MS8.

- NaOH will react with ethanoic acid to form sodium ethanoate and water (1 mark)

- Shift to the right (favors CH3COOH dissociation) to replace the ethanoic acid that is consumed (1 mark)

- Le Chatelier's principle: NaOH is a base, so adding it will reduce the concentration of H+ ions (1 mark)

MS9.

- pH = -log[H+] (1 mark)

- Ka = [H+][CH3COO-] / [CH3COOH] (1 mark)

- [H+] = Ka[CH3COOH] / [CH3COO-] = (1.8 × 10^-5) (0.1) / (0.1) = 1.8 × 10^-5 (1 mark)

- pH = -log(1.8 × 10^-5) = 4.74 (1 mark)

MS10.

- State of equilibrium where the concentration of a solute in a saturated solution remains constant. (1 mark)

- Occurs when the rate of dissolution equals the rate of crystallization. (1 mark)