AP Chemistry Quiz: Enthalpy Of Formation

20 questions · exam conditions

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Enthalpy Of FormationQuestion 1 of 20

Use the standard enthalpies of formation, $\Delta H_f^\circ$ , to calculate $\Delta H_{\text{rxn}}^\circ$ for the reaction at 298 K:\n\n $\text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g)$ \n\nFormation data (kJ/mol): $\Delta H_f^\circ[\text{H}_2(g)] = 0$ , $\Delta H_f^\circ[\text{Cl}_2(g)] = 0$ , $\Delta H_f^\circ[\text{HCl}(g)] = -92$ .

+184 kJ/mol

-92 kJ/mol

-184 kJ/mol

+92 kJ/mol

0 kJ/mol

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AP Chemistry Quiz

AP Chemistry Quiz: Enthalpy Of Formation

Practice Enthalpy Of Formation in AP Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Enthalpy Of Formation, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

+184 kJ/mol
-92 kJ/mol
-184 kJ/mol (correct answer)
+92 kJ/mol
0 kJ/mol

Explanation: This question tests the skill of calculating the standard enthalpy change of a reaction using standard enthalpies of formation. Using Hess's law, $\Delta H_{\text{rxn}}^\circ$ = $\Sigma$ (coefficients × $\Delta H_f^\circ$ products) - $\Sigma$ (coefficients × $\Delta H_f^\circ$ reactants). For this, products 2 HCl at -92 kJ/mol sum to -184 kJ/mol, reactants H2 and Cl2 both 0, so -184 - 0 = -184 kJ/mol, exothermic. This reflects the energy release in forming HCl bonds. A tempting distractor is -92 kJ/mol, due to the misconception of omitting the coefficient 2 for HCl. A key strategy is to list all terms with coefficients before summing to ensure no multiplication errors in enthalpy calculations.

Question 2

Use the standard enthalpies of formation, $\Delta H_f^\circ$ , given below to calculate $\Delta H_{\text{rxn}}^\circ$ for the reaction at 298 K:

$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$

Formation data (kJ/mol): $\Delta H_f^\circ[\text{CH}_4(g)] = -75$ , $\Delta H_f^\circ[\text{O}_2(g)] = 0$ , $\Delta H_f^\circ[\text{CO}_2(g)] = -394$ , $\Delta H_f^\circ[\text{H}_2\text{O}(l)] = -286$ .

-891 kJ/mol (correct answer)
+891 kJ/mol
-605 kJ/mol
-103 kJ/mol
-966 kJ/mol

Explanation: This question tests the skill of calculating the standard enthalpy change of a reaction using standard enthalpies of formation. The standard enthalpy of reaction, $\Delta H_{\text{rxn}}^\circ$ , is found by subtracting the sum of the standard enthalpies of formation of the reactants, each multiplied by their stoichiometric coefficients, from that of the products. For this reaction, the products' enthalpies sum to $-394 \text{ kJ/mol}$ for CO2 plus 2 times $-286 \text{ kJ/mol}$ for H2O(l), equaling $-966 \text{ kJ/mol}$ , while the reactants sum to $-75 \text{ kJ/mol}$ for CH4 and 0 for O2. Subtracting gives $-966 - (-75) = -891 \text{ kJ/mol}$ , indicating an exothermic reaction. A tempting distractor is $-605 \text{ kJ/mol}$ , which arises from the misconception of forgetting to multiply the enthalpy of formation of H2O by its coefficient of 2. Always remember to multiply each enthalpy of formation by the stoichiometric coefficient and ensure the states of matter match the given data when calculating $\Delta H_{\text{rxn}}^\circ$ .

Question 3

Use the standard enthalpies of formation, $\Delta H_f^\circ$ , given to calculate $\Delta H_{rxn}^\circ$ for the reaction below.

$2\text{H}_2\text{O}(l) \rightarrow 2\text{H}_2(g) + \text{O}_2(g)$

Given: $\Delta H_f^\circ[\text{H}_2\text{O}(l)] = -286\ \text{kJ/mol}$ , $\Delta H_f^\circ[\text{H}_2(g)] = 0\ \text{kJ/mol}$ , $\Delta H_f^\circ[\text{O}_2(g)] = 0\ \text{kJ/mol}$ .

-286 kJ/mol
+572 kJ/mol (correct answer)
+286 kJ/mol
-572 kJ/mol
+858 kJ/mol

Explanation: This question tests your ability to calculate the standard enthalpy of reaction using standard enthalpies of formation. This is the reverse of water formation, so we apply $\Delta H^\circ_{\text{rxn}} = \Sigma (\Delta H^\circ_f \text{ products}) - \Sigma (\Delta H^\circ_f \text{ reactants})$ . For products: $2\text{H}_2(g)$ and $\text{O}_2(g)$ are elements in standard states with $\Delta H^\circ_f = 0$ kJ/mol each, giving a total of 0 kJ/mol. For reactants: $2\text{H}_2\text{O}(l)$ has $\Delta H^\circ_f = 2(-286) = -572$ kJ/mol. Therefore, $\Delta H^\circ_{\text{rxn}} = 0 - (-572) = +572$ kJ/mol. A common mistake (choice C) is using the $\Delta H^\circ_f$ for only 1 mole of water, giving +286 kJ/mol. The positive value confirms that decomposing water requires energy input, as expected for breaking stable bonds.

Question 4

Given the following $\Delta H_f^\circ$ values at 298 K: $\Delta H_f^\circ[\mathrm{H_2O(l)}]=-286\ \mathrm{kJ\,mol^{-1}}$ , $\Delta H_f^\circ[\mathrm{H_2O(g)}]=-242\ \mathrm{kJ\,mol^{-1}}$ . What is $\Delta H_{\mathrm{rxn}}^\circ$ for $\mathrm{H_2O(l)\rightarrow H_2O(g)}$ ?

0 kJ/mol
+528 kJ/mol
-528 kJ/mol
-44 kJ/mol
+44 kJ/mol (correct answer)

Explanation: This question tests the skill of calculating enthalpy of reaction using standard enthalpies of formation. For water vaporization: ΔH°rxn = [1(-242)] - [1(-286)] = -242 - (-286) = -242 + 286 = +44 kJ/mol. The positive value confirms that vaporization is endothermic, requiring energy input. A common mistake would be to calculate -44 kJ/mol (choice B) by subtracting in the wrong order, confusing the direction of the phase change. Remember that ΔH°rxn = products minus reactants, and vaporization always requires energy input, so ΔH must be positive.