41. According to Hess’s Law, if a reaction can be expressed as a series of steps, then:

• A) The enthalpy change for the overall reaction is the sum of the enthalpy changes of each step
• B) The enthalpy change is only applicable at constant pressure
• C) The reaction must occur in one step
• D) The heat of reaction cannot be determined
• Answer: A

42. What type of reaction has a ΔH value that is greater than zero?

• A) Exothermic
• B) Endothermic
• C) Isothermal
• D) Isochoric
• Answer: B

43. The change in internal energy is affected by:

• A) The type of reaction
• B) The path of the reaction
• C) The amount of work done
• D) Both A and C
• Answer: D

44. The heat of reaction can be calculated using:

• A) The bond energies of reactants and products
• B) The temperature change only
• C) The concentration of reactants
• D) The volume of reactants
• Answer: A

45. Which of the following conditions defines a reversible process?

• A) It occurs rapidly
• B) It is at equilibrium at every stage
• C) It is an isothermal process
• D) It involves a net increase in entropy
• Answer: B

46. What happens to the internal energy of a system when it absorbs heat and does work on the surroundings?

• A) Increases
• B) Decreases
• C) Remains constant
• D) Cannot be determined
• Answer: B

47. Which of the following is NOT a consequence of the First Law of Thermodynamics?

• A) Energy can be created from nothing
• B) Energy can be transformed from one form to another
• C) The total energy of an isolated system is constant
• D) Heat can be converted to work
• Answer: A

48. In an isobaric process, which of the following is true?

• A) The pressure remains constant
• B) The volume remains constant
• C) The temperature remains constant
• D) The system is isolated
• Answer: A

49. Which of the following equations is used to calculate the heat (q) absorbed or released during a temperature change?

• A) q = m × C × ΔT
• B) q = n × ΔH
• C) q = W + Q
• D) q = ΔU + PΔV
• Answer: A

50. The relationship between ΔH, ΔU, and the work done (W) can be expressed as:

• A) ΔH = ΔU + W
• B) ΔH = ΔU – W
• C) ΔH = ΔU + Q
• D) ΔH = ΔU + PΔV
• Answer: D