OAT OAT: General Chemistry Questions and Answers — Questions and Answers
Question 1: The rate law for a particular reaction is determined to be: Rate = k[A][B]². If the concentration of reactant A is doubled and the concentration of reactant B is halved, what will be the effect on the initial rate of the reaction?
- The rate will be quadrupled.
- The rate will be doubled.
- The rate will remain the same.
- The rate will be halved. (Correct answer)
Correct answer: The rate will be halved.
The rate's dependence on reactant concentration is given by the rate law. The reaction is first order with respect to [A] and second order with respect to [B]. Let the initial rate be R1 = k[A][B]². The new rate, R2, will be k[2A][(1/2)B]² = k(2[A])(1/4[B]²) = (2)(1/4) * k[A][B]² = (1/2)R1. Therefore, the rate is halved.
Question 2: During the titration of 25.0 mL of 0.100 M HCl with 0.100 M NaOH, what is the pH of the solution at the equivalence point at 25°C?
- 3.00
- 7.00 (Correct answer)
- 5.00
- 9.00
Correct answer: 7.00
The equivalence point in the titration of a strong acid (HCl) with a strong base (NaOH) results in a solution containing only a neutral salt (NaCl) and water. Since NaCl is the salt of a strong acid and a strong base, neither the Na⁺ nor the Cl⁻ ions hydrolyze water to produce H⁺ or OH⁻. Consequently, the solution is neutral with a pH of 7.00 at 25°C.
Question 3: Which of the following sets of quantum numbers (n, l, m_l, m_s) is NOT permissible for an electron in an atom?
- (2, 1, -1, +1/2)
- (3, 2, 0, -1/2)
- (4, 3, -4, +1/2) (Correct answer)
- (5, 0, 0, +1/2)
Correct answer: (4, 3, -4, +1/2)
The rules for quantum numbers are: n must be a positive integer (1, 2, 3...); l can range from 0 to n-1; m_l can range from -l to +l, including 0. In choice C, n=4 and l=3, which is allowed. However, m_l is given as -4. For l=3, the allowed values of m_l are -3, -2, -1, 0, +1, +2, and +3. An m_l value of -4 is therefore not permissible.
Question 4: A chemical reaction is found to have a negative change in enthalpy (ΔH < 0) and a positive change in entropy (ΔS > 0). Which of the following statements is true regarding this reaction?
- The reaction is spontaneous at all temperatures. (Correct answer)
- The reaction is spontaneous only at high temperatures.
- The reaction is spontaneous only at low temperatures.
- The reaction is non-spontaneous at all temperatures.
Correct answer: The reaction is spontaneous at all temperatures.
The spontaneity of a reaction is determined by the Gibbs free energy change, ΔG = ΔH - TΔS. A reaction is spontaneous if ΔG is negative. Given that ΔH is negative and ΔS is positive, the term -TΔS will always be negative (since temperature, T, in Kelvin is always positive). Therefore, ΔG is the sum of a negative term (ΔH) and another negative term (-TΔS), which guarantees that ΔG will be negative regardless of the temperature.
Question 5: Assuming ideal behavior, which of the following aqueous solutions would be expected to have the lowest freezing point?
- 0.10 m NaCl
- 0.10 m C₆H₁₂O₆ (glucose)
- 0.08 m MgCl₂ (Correct answer)
- 0.10 m KBr
Correct answer: 0.08 m MgCl₂
Freezing point depression is a colligative property that depends on the effective molality of solute particles (molality × van 't Hoff factor, i). The solution with the highest effective molality will have the largest freezing point depression and thus the lowest freezing point. Let's calculate the effective molality for each: A) NaCl (i=2): 0.10 m * 2 = 0.20 m. B) Glucose (i=1): 0.10 m * 1 = 0.10 m. C) MgCl₂ (i=3): 0.08 m * 3 = 0.24 m. D) KBr (i=2): 0.10 m * 2 = 0.20 m. Since 0.08 m MgCl₂ has the highest effective molality (0.24 m), it will experience the greatest freezing point depression.
Question 6: According to the reaction 2 NaN₃(s) → 2 Na(s) + 3 N₂(g), how many liters of nitrogen gas, measured at Standard Temperature and Pressure (STP), are produced from the complete decomposition of 130.0 g of sodium azide (NaN₃)? (Molar mass of NaN₃ = 65.0 g/mol)
- 22.4 L
- 44.8 L
- 67.2 L (Correct answer)
- 89.6 L
Correct answer: 67.2 L
First, calculate the moles of NaN₃: (130.0 g) / (65.0 g/mol) = 2.00 mol NaN₃. Next, use the stoichiometric ratio from the balanced equation to find the moles of N₂ produced: (2.00 mol NaN₃) × (3 mol N₂ / 2 mol NaN₃) = 3.00 mol N₂. At STP (0°C and 1 atm), one mole of any ideal gas occupies 22.4 liters. Finally, calculate the volume of N₂ gas: (3.00 mol N₂) × (22.4 L/mol) = 67.2 L.
OAT OAT: General Chemistry Questions and Answers
The rate law for a particular reaction is determined to be: Rate = k[A][B]².
If the concentration of reactant A is doubled and the concentration of reactant B is halved, what will be the effect on the initial rate of the reaction?