AP Chemistry Practice Quiz: Hess’s Law
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 14 questions to check your progress.
Question 1 of 14
All Questions (14)
A) The sign of ΔH is reversed.
B) The value of ΔH is squared.
C) The value of ΔH is halved.
D) The value of ΔH remains unchanged.
Correct Answer: A
One of the fundamental principles of Hess's Law is that reversing a reaction reverses the sign of its ΔH. For example, if the forward reaction is endothermic (+ΔH), the reverse reaction will be exothermic (-ΔH).
A) -100 kJ
B) -50 kJ
C) -200 kJ
D) +200 kJ
Correct Answer: C
Hess's Law states that if a reaction is multiplied by a factor 'c', the ΔH is also multiplied by 'c'. In this case, the reaction is multiplied by 2, so the ΔH is -100 kJ * 2 = -200 kJ.
A) The law of definite proportions
B) The law of conservation of mass
C) The law of conservation of energy
D) The ideal gas law
Correct Answer: C
The provided content states that due to energy conservation, the net enthalpy change of an overall process is equal to the sum of the enthalpy changes of its individual steps. This directly links Hess's Law to the law of conservation of energy.
A) +85 kJ
B) -35 kJ
C) +35 kJ
D) -85 kJ
Correct Answer: B
According to Hess's Law, the enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps. Therefore, ΔH_total = ΔH₁ + ΔH₂ = (+25 kJ) + (-60 kJ) = -35 kJ.
A) +20 kJ
B) -20 kJ
C) -120 kJ
D) +120 kJ
Correct Answer: C
To find the enthalpy change for the overall reaction, we sum the individual reactions and their ΔH values. (A + B → C) + (C + B → D) gives A + 2B → D. The net ΔH is the sum of the individual enthalpies: (-50 kJ) + (-70 kJ) = -120 kJ.
A) -4580 kJ
B) +1300 kJ
C) -1300 kJ
D) +4580 kJ
Correct Answer: C
To get the target reaction, we need P₄O₆ on the reactant side, so we must reverse reaction 1. This changes its ΔH to +1640 kJ. We keep reaction 2 as is. Summing the manipulated reactions: (P₄O₆(s) → P₄(s) + 3O₂(g)) + (P₄(s) + 5O₂(g) → P₄O₁₀(s)) gives the target equation. The net ΔH is (+1640 kJ) + (-2940 kJ) = -1300 kJ.
A) The overall enthalpy is the product of the step enthalpies.
B) The overall enthalpy is the average of the step enthalpies.
C) The overall enthalpy is the sum of the step enthalpies.
D) The overall enthalpy is independent of the step enthalpies.
Correct Answer: C
The content explicitly states that the net enthalpy change of an overall process at constant pressure is equal to the sum of the enthalpy changes of its individual steps. This is the core concept of Hess's Law.
A) +20 kJ
B) +65 kJ
C) +35 kJ
D) -10 kJ
Correct Answer: B
We need to arrange the given equations to sum to the target equation. Keep reaction 1 as is: M + N → Q, ΔH = +50 kJ. To get 0.5P on the product side, we must reverse reaction 2 and multiply it by 0.5. Reversing gives: 2Q → P, ΔH = +30 kJ. Multiplying by 0.5 gives: Q → 0.5P, ΔH = +15 kJ. Now, sum the modified reactions: (M + N → Q) + (Q → 0.5P) = M + N → 0.5P. The total ΔH is the sum of the modified enthalpies: (+50 kJ) + (+15 kJ) = +65 kJ.
A) By multiplying the enthalpy changes of the three steps.
B) By summing the enthalpy changes of the three steps.
C) By taking the enthalpy change of the final step only.
D) By averaging the enthalpy changes of the three steps.
Correct Answer: B
Hess's Law states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step in the reaction, regardless of how many steps there are. This is a direct application of the principle of energy conservation.
A) +180 kJ
B) -180 kJ
C) +90 kJ
D) -90 kJ
Correct Answer: D
To obtain the target reaction from the given reaction, two manipulations are needed. First, the reaction must be reversed, which changes the sign of ΔH from +180 kJ to -180 kJ. Second, the reversed reaction (2Z → 2X + Y) must be multiplied by 0.5 (or divided by 2) to get Z on the reactant side with a coefficient of 1. This means the new ΔH must also be multiplied by 0.5: -180 kJ * 0.5 = -90 kJ.
A) +60 kJ
B) +240 kJ
C) -60 kJ
D) -240 kJ
Correct Answer: A
The goal is to combine the steps to form the target reaction. Reaction 1 has A and D on the correct sides, so we keep it as is. Reaction 2 has B on the correct side, so we also keep it as is. Summing the two reactions: (A + 2C → D) + (B → 2C) gives A + B + 2C → D + 2C. The intermediate '2C' cancels from both sides, leaving the target reaction A + B → D. The net enthalpy change is the sum of the individual ΔH values: (+150 kJ) + (-90 kJ) = +60 kJ.
A) ΔH_total = ΔH₁ × ΔH₂ × ΔH₃
B) ΔH_total = (ΔH₁ + ΔH₂ + ΔH₃) / 3
C) ΔH_total = ΔH₁ + ΔH₂ + ΔH₃
D) ΔH_total is unrelated to the steps in Pathway 2.
Correct Answer: C
Hess's Law is based on the concept that enthalpy is a state function. This means the total enthalpy change of a process depends only on the initial and final states, not on the path taken. Therefore, the enthalpy change of the direct path (Pathway 1) must be equal to the sum of the enthalpy changes of the steps in the indirect path (Pathway 2).
A) -137 kJ
B) -6514 kJ
C) +137 kJ
D) -274 kJ
Correct Answer: A
To find the ΔH for the target reaction, manipulate the given equations. 1) Multiply Eq. 1 by 1/2: C₂H₄ + 3O₂ → 2CO₂ + 2H₂O, ΔH = -1411 kJ. 2) Reverse and multiply Eq. 2 by 1/2: 2CO₂ + 3H₂O → C₂H₆ + 3.5O₂, ΔH = +1560 kJ. 3) Multiply Eq. 3 by 1/2: H₂ + 0.5O₂ → H₂O, ΔH = -286 kJ. Summing these three new equations and their ΔH values, the O₂, CO₂, and H₂O cancel out, leaving the target reaction. The total ΔH = (-1411) + (+1560) + (-286) = -137 kJ.
A) -80 kJ
B) +320 kJ
C) +20 kJ
D) -220 kJ
Correct Answer: D
To obtain the target reaction (2A + 3B → 2D), we must manipulate the given steps. First, take equation 1 and multiply it by 1/2 to get the correct coefficient for A and D: (1/2) * (4A + 2C → 4D) becomes 2A + C → 2D, and the ΔH becomes (-200 kJ) * 1/2 = -100 kJ. Next, to get 3B as a reactant, we must reverse equation 2: (C → 3B) becomes 3B → C, and the ΔH becomes -120 kJ. Now, sum the two new equations: (2A + C → 2D) + (3B → C). The intermediate 'C' appears on both the reactant and product side, so it cancels out. The resulting equation is 2A + 3B → 2D. The net ΔH is the sum of the ΔH values from the manipulated steps: (-100 kJ) + (-120 kJ) = -220 kJ.