Q.Given $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$; $\Delta_r H = -92.4$ kJ mol$^{-1}$. What is the standard enthalpy of formation of $NH_3$ gas?

Showing the 12 most recent of 30 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The $\Delta_f H^{\ominus}$ of $\mathrm{AO(g)}$, $\mathrm{AO_2(g)}$, $\mathrm{B_2O(g)}$ and $\mathrm{B_2O_4(g)}$ is respectively $-110, -393, +81$ and $+9.7 \; \mathrm{kJ\,mol^{-1}}$. What is $\Delta_r H^{\ominus}$ (in $\mathrm{kJ\,mol^{-1}}$) of the following reaction at 298 K? $\mathrm{B_2O_4(g) + 3\,AO(g) \rightarrow B_2O(g) + 3\,AO_2(g)}$ (A) $-777.7$ (B) $-1179$ (C) $+777.7$ (D) $+1179$

   ›Reveal solutionSolution

   This tests Hess's law applied to a reaction written as a linear combination of formation reactions. $\Delta_r H^{\ominus} = -777.7\ \mathrm{kJ\,mol^{-1}}$.

   Concept and Intuition

   Any reaction's standard enthalpy can be built from the standard enthalpies of formation of every species, because $\Delta H$ is a state function — it doesn't matter what path you imagine to reach the products, only where you start and end. For a reaction $a\,W \rightarrow b\,X$, that path is: decompose the reactants into their elements (undoing formation, so with a $-$ sign), then reform the products from elements (using formation values with a $+$ sign):

   $\Delta_r H^{\ominus} = \sum_{\text{products}} \nu\, \Delta_f H^{\ominus} - \sum_{\text{reactants}} \nu\, \Delta_f H^{\ominus}$

   Step-by-Step Solution

   1. Write the reaction: $\mathrm{B_2O_4(g) + 3\,AO(g) \rightarrow B_2O(g) + 3\,AO_2(g)}$.
   2. List the given formation enthalpies: $\Delta_fH^\ominus[\mathrm{AO}] = -110$, $\Delta_fH^\ominus[\mathrm{AO_2}] = -393$, $\Delta_fH^\ominus[\mathrm{B_2O}] = +81$, $\Delta_fH^\ominus[\mathrm{B_2O_4}] = +9.7$ (all $\mathrm{kJ\,mol^{-1}}$).
   3. Sum for products: $\Delta_fH^\ominus[\mathrm{B_2O}] + 3\Delta_fH^\ominus[\mathrm{AO_2}] = 81 + 3(-393) = 81 - 1179 = -1098\ \mathrm{kJ}$. …

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2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.At T(K), enthalpy of combustion of C(s), $\mathrm{H_2(g)}$ and $\mathrm{CH_4(g)}$ are $-393.5$, $-286$, $-890$ kJ $\mathrm{mol^{-1}}$ respectively. What is the enthalpy of formation of methane (in kJ $\mathrm{mol^{-1}}$)? (A) 54.6 (B) -75.5 (C) 99.7 (D) 89.6

   ›Reveal solutionSolution

   Hess's-law combination of the three combustion enthalpies (C, H$_2$, CH$_4$) gives the formation enthalpy of methane as $-75.5$ kJ/mol.

   Concept and Intuition

   The enthalpy of formation of methane, $\mathrm{C(s)+2H_2(g)\to CH_4(g)}$, cannot be measured directly (carbon doesn't cleanly react with hydrogen to give pure methane in a calorimeter), so it is obtained indirectly via Hess's law using combustion data, since combustion enthalpies of all three substances (C, H$_2$, CH$_4$) are easy to measure directly by burning them in oxygen.

   Step-by-Step Solution

   1. Write the three given combustion reactions:
      • $\mathrm{C(s)+O_2(g)\to CO_2(g)}$, $\Delta H_1=-393.5$ kJ/mol
      • $\mathrm{H_2(g)+\tfrac12O_2(g)\to H_2O(l)}$, $\Delta H_2=-286$ kJ/mol
      • $\mathrm{CH_4(g)+2O_2(g)\to CO_2(g)+2H_2O(l)}$, $\Delta H_3=-890$ kJ/mol
   2. Target reaction: $\mathrm{C(s)+2H_2(g)\to CH_4(g)}$, $\Delta H_f=?$ …

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3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.Combustion of methane gives $CO_2(g)$ and $H_2O(l)$. What is enthalpy of combustion ($\Delta_cH^\ominus$ in kJ mol$^{-1}$) of $CH_4(g)$ at 298 K? ($\Delta_fH^\ominus(CH_4(g)) = -x$ kJ mol$^{-1}$; $\Delta_fH^\ominus(CO_2(g)) = -y$ kJ mol$^{-1}$; $\Delta_fH^\ominus(H_2O(l)) = -z$ kJ mol$^{-1}$) (A) $-(y+2z-x)$ (B) $(2z-y+x)$ (C) $(2z+x-y)$ (D) $-(2y+z+x)$

   ›Reveal solutionSolution

   Enthalpy of reaction = sum of $\Delta_fH^\ominus$(products) − sum of $\Delta_fH^\ominus$(reactants), remembering $O_2(g)$ (an element in its standard state) contributes zero. The result simplifies to $-(y+2z-x)$.

   Concept and Intuition

   Hess's law lets us compute a reaction's enthalpy purely from tabulated standard enthalpies of formation, since enthalpy is a state function and doesn't depend on the path taken. Elements in their standard reference state (like $O_2(g)$ here) are defined to have $\Delta_fH^\ominus=0$, since "forming" an element from itself involves no change.

   Step-by-Step Solution

   1. Write the balanced combustion equation: $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$.
   2. Apply Hess's law: $\Delta_cH^\ominus=\left[\Delta_fH^\ominus(CO_2)+2\,\Delta_fH^\ominus(H_2O)\right]-\left[\Delta_fH^\ominus(CH_4)+2\,\Delta_fH^\ominus(O_2)\right]$. …

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4. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.For the reaction, $2Al_2O_3(s) \to 4Al(s) + 3O_2(g)$, $\Delta H = 3340\ kJ$. What is the enthalpy of formation of $Al_2O_3(s)$ (in $kJ$)? (A) $+1670$ (B) $-3340$ (C) $+3340$ (D) $-1670$

   ›Reveal solutionSolution

   This tests scaling a given reaction enthalpy to a per-mole basis and then reversing it to get the enthalpy of formation. The enthalpy of formation of $Al_2O_3(s)$ is $-1670\ kJ$.

   Concept and Intuition

   Enthalpy of formation refers strictly to forming one mole of compound from its elements in their standard states. Given data for a different stoichiometry (here, decomposition of 2 mol), you must first scale to 1 mole, and then flip the sign if the given reaction runs the opposite direction (decomposition vs. formation) — by Hess's law, $\Delta H_{decomposition} = -\Delta H_{formation}$.

   Step-by-Step Solution

   1. Given: $2Al_2O_3(s) \to 4Al(s) + 3O_2(g)$, $\Delta H = +3340\ kJ$ (this is decomposition of 2 mol $Al_2O_3$, and it's endothermic as written since forming the stable oxide releases energy, so breaking it apart consumes energy). …

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5. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.What is the bond enthalpy (in $kJ\ mol^{-1}$) of C-H in ethane? ($\Delta_f H^{\ominus}(\mathrm{C_2H_6(g)}) = -85\ kJ\ mol^{-1}$; $\mathrm{H_2(g) \to 2H(g)};\ \Delta_a H^{\ominus} = 435\ kJ\ mol^{-1}$; $\mathrm{C(s) \to C(g)};\ \Delta_a H^{\ominus} = 715\ kJ\ mol^{-1}$ $\Delta_a H^{\ominus}(\mathrm{C-C}) = 347\ kJ\ mol^{-1}$) (A) 412.2 (B) 402.8 (C) 390.7 (D) 380.6

   ›Reveal solutionSolution

   Ethane's formation enthalpy is decomposed into atomisation energies of C and H₂ minus the total bond energy released forming all its C–C and C–H bonds; solving for the unknown C–H bond enthalpy gives the answer.

   Concept and Intuition

   Think of forming $C_2H_6(g)$ in two steps: first atomise 2 mol C(s) and 3 mol H₂(g) into free gaseous atoms (costs energy, the atomisation enthalpies), then let those atoms bond together into one ethane molecule (releases energy equal to the sum of all bond enthalpies formed: one C–C bond and six C–H bonds). The net of these two steps must equal the known $\Delta_fH$ of ethane — Hess's law via a bond-enthalpy cycle.

   Step-by-Step Solution

   1. Atomisation step (energy absorbed): $2\times\Delta_aH(C)+3\times\Delta_aH(H_2)=2(715)+3(435)=1430+1305=2735$ kJ/mol.
   2. Bond-formation step (energy released) forming $C_2H_6$ from atoms: $\Delta_aH(C\text{-}C)+6\times BE(C\text{-}H)=347+6x$. …

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6. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.What is $\Delta_r H^{\ominus}$ (in $kJ\ mol^{-1}$) for the following reaction at 298 K? $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$ (Given: $\Delta_f H^{\ominus}$ of $C_3H_8(g), CO_2(g)$ and $H_2O(l)$ is $-104, -393$ and $-285\ kJ\ mol^{-1}$ respectively) (A) +2215 (B) -2215 (C) -2427 (D) -2323

   ›Reveal solutionSolution

   A standard Hess's-law combustion enthalpy calculation using given formation enthalpies gives $\Delta_rH^{\ominus}=-2215$ kJ/mol.

   Concept and Intuition

   The standard enthalpy of a reaction can always be obtained from the standard enthalpies of formation of reactants and products, since enthalpy is a state function (Hess's law): $\Delta_rH^{\ominus} = \sum n_p\Delta_fH^{\ominus}(\text{products}) - \sum n_r\Delta_fH^{\ominus}(\text{reactants})$. Elements in their standard states (like $O_2(g)$ here) have $\Delta_fH^{\ominus}=0$ by definition.

   Step-by-Step Solution

   1. Balanced reaction: $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$.
   2. Sum of products' formation enthalpies: $3(-393) + 4(-285) = -1179 + (-1140) = -2319$ kJ.
   3. Sum of reactants' formation enthalpies: $(-104) + 5(0) = -104$ kJ.
   4. $\Delta_rH^{\ominus} = (-2319) - (-104) = -2319 + 104 = -2215$ kJ/mol. …

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7. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.Consider the following reaction $CaCO_3(s) \rightarrow CaO(s) + CO_2(g) - 178\ kJ$. The standard enthalpy of formation of $CaCO_3(s)$ and $CO_2(g)$ is $-1207$ and $-393\ kJ\,mol^{-1}$ respectively. What is $\Delta_f H^{\ominus}$ (in $kJ\,mol^{-1}$) of $CaO(s)$? (A) $-636$ (B) $+636$ (C) $-814$ (D) $+814$

   ›Reveal solutionSolution

   Applying Hess's law to the calcination reaction (an endothermic decomposition, $\Delta_rH=+178$ kJ) and the given $\Delta_fH$ values for $CaCO_3$ and $CO_2$ gives $\Delta_fH(CaO)=-636$ kJ/mol.

   Concept and Intuition

   The decomposition of limestone, $CaCO_3(s)\to CaO(s)+CO_2(g)$, is a well-known endothermic reaction (calcination requires heating). When such an equation is written with a heat term subtracted on the product side (an older textbook convention: "reactants $\to$ products $-\ Q$" means heat $Q$ is absorbed, i.e. $\Delta_rH=+Q$), it is equivalent to the modern convention $\Delta_rH=+178$ kJ/mol. Hess's law then lets us find any one enthalpy of formation if the others and the reaction enthalpy are known.

   Step-by-Step Solution

   1. Reaction: $CaCO_3(s)\to CaO(s)+CO_2(g)$, with $\Delta_rH=+178$ kJ/mol (endothermic, matching the physical process).
   2. Hess's law: $\Delta_rH=\Delta_fH(CaO)+\Delta_fH(CO_2)-\Delta_fH(CaCO_3)$.
   3. Substitute known values: $178=\Delta_fH(CaO)+(-393)-(-1207)$.
   4. Simplify: $178=\Delta_fH(CaO)-393+1207=\Delta_fH(CaO)+814$. …

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8. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.Consider the following reaction $CCl_4(g) \rightarrow C(g) + 4\,Cl(g) - 1304\ kJ$. What is $\Delta_{vap} H^{\ominus}$ of $CCl_4(l)$ (in $kJ\,mol^{-1}$)? [Given: $\Delta_f H^{\ominus}(CCl_4(l)) = -135.5\ kJ\,mol^{-1}$, $\Delta_a H^{\ominus}(C) = 715\ kJ\,mol^{-1}$, $\Delta_a H^{\ominus}(Cl_2) = 242\ kJ\,mol^{-1}$] (A) $+272.5$ (B) $-30.5$ (C) $-272.5$ (D) $+30.5$

   ›Reveal solutionSolution

   A Hess's-law thermochemical cycle linking $CCl_4(l)$, $CCl_4(g)$, and the fully atomized elements gives $\Delta_{vap}H(CCl_4)=+30.5$ kJ/mol — consistent with the known real vaporization enthalpy of carbon tetrachloride (~30 kJ/mol).

   Concept and Intuition

   We want $\Delta_{vap}H$ for $CCl_4(l)\to CCl_4(g)$. We're given the atomization of gaseous $CCl_4$ ($CCl_4(g)\to C(g)+4Cl(g)$, endothermic, physically $\Delta H=+1304$ kJ/mol — bond-breaking always costs energy, so the "$-1304$" in the problem uses the same "heat absorbed" notation as the calcination reaction), the formation enthalpy of liquid $CCl_4$, and atomization enthalpies of carbon (graphite $\to$ gas) and of $Cl_2$ (molecule $\to$ 2 atoms). We build two equivalent thermochemical paths from graphite + chlorine gas to fully atomized carbon and chlorine, and equate them.

   Step-by-Step Solution

   1. Target unknown: $CCl_4(l)\xrightarrow{\Delta_{vap}H} CCl_4(g)$.
   2. Given atomization of gaseous $CCl_4$: $CCl_4(g)\to C(g)+4Cl(g)$, $\Delta H=+1304$ kJ/mol (endothermic — breaking 4 C–Cl bonds).
   3. Given: $\Delta_fH(CCl_4,l)=-135.5$ kJ/mol, i.e. $C(\text{gr})+2Cl_2(g)\to CCl_4(l)$, $\Delta H=-135.5$.
   4. Given atomization enthalpies: $C(\text{gr})\to C(g)$, $\Delta_aH=+715$ kJ/mol; $Cl_2(g)\to 2Cl(g)$, $\Delta_aH=+242$ kJ/mol (per mole $Cl_2$), so for 2 mol $Cl_2\to4Cl(g)$: $2\times242=484$ kJ.
   5. Direct path (graphite + $Cl_2$ straight to atoms): $C(\text{gr})+2Cl_2(g)\to C(g)+4Cl(g)$: $\Delta H=715+484=1199$ kJ. …

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9. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.One mole of $\mathrm{C_2H_5OH(l)}$ was completely burnt in oxygen to form $\mathrm{CO_2(g)}$ and $\mathrm{H_2O(l)}$. The standard enthalpy of formation ($\Delta_f H^{\ominus}$) of $\mathrm{C_2H_5OH(l)}$, $\mathrm{CO_2(g)}$ and $\mathrm{H_2O(l)}$ is $x, y, z$ kJ $\mathrm{mol^{-1}}$ respectively. What is $\Delta_r H^{\ominus}$ (in kJ $\mathrm{mol^{-1}}$) for this reaction? (A) $(2y + 3z + x)$ (B) $(2y - 3z + x)$ (C) $(x - 2y - 3z)$ (D) $(2y + 3z - x)$

   ›Reveal solutionSolution

   This tests Hess's law / the standard formula $\Delta_rH^\ominus = \sum \Delta_fH^\ominus(\text{products}) - \sum \Delta_fH^\ominus(\text{reactants})$ for ethanol combustion, giving $2y+3z-x$.

   Concept and Intuition

   Any reaction's standard enthalpy can be built from the standard enthalpies of formation of all species involved, since enthalpy is a state function; elements in their standard reference state (like $O_2(g)$) contribute zero to this sum by definition.

   Step-by-Step Solution

   1. Write the balanced combustion equation: $\mathrm{C_2H_5OH(l) + 3O_2(g) \longrightarrow 2CO_2(g) + 3H_2O(l)}$.
   2. Apply Hess's law: $\Delta_rH^\ominus = \left[2\,\Delta_fH^\ominus(CO_2) + 3\,\Delta_fH^\ominus(H_2O)\right] - \left[\Delta_fH^\ominus(C_2H_5OH) + 3\,\Delta_fH^\ominus(O_2)\right]$. …

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10. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
    Q.The standard enthalpy of formation of $\mathrm{C_2H_4(g)}$, $\mathrm{CO_2(g)}$ and $\mathrm{H_2O(\ell)}$ are 52, -394 and -286 kJ mol$^{-1}$ respectively. The heat evolved by burning 7 g of $\mathrm{C_2H_4(g)}$ is (A) 353 kJ (B) 1412 kJ (C) 706 kJ (D) 1306 kJ

    ›Reveal solutionSolution

    Hess's law gives $\Delta H_{comb}=-1412\ \text{kJ/mol}$ for C$_2$H$_4$; scaling to 7 g (a quarter mole) gives 353 kJ evolved.

    Concept and Intuition

    The enthalpy of combustion can be computed from standard enthalpies of formation via $\Delta H_{rxn} = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$ Once we know the heat released per mole of fuel burned, we scale it to the actual mass burned using the molar mass.

    Step-by-Step Solution

    1. Balanced combustion equation: $\mathrm{C_2H_4(g) + 3O_2(g) \to 2CO_2(g) + 2H_2O(\ell)}$
    2. Apply Hess's law: $\Delta H_{comb} = [2\Delta H_f(CO_2) + 2\Delta H_f(H_2O)] - [\Delta H_f(C_2H_4) + 3\Delta H_f(O_2)]$
    3. $\Delta H_f(O_2)=0$ (element in standard state). Substituting values: $\Delta H_{comb} = [2(-394)+2(-286)] - [52+0] = [-788-572]-52 = -1360-52=-1412\ \text{kJ/mol}$
    4. Molar mass of C$_2$H$_4$ $= 2(12)+4(1)=28\ \text{g/mol}$. Moles in 7 g: $n = \frac{7}{28} = 0.25\ \text{mol}$ …

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11. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
    Q.The value of $\Delta_rH$ for the reaction $\mathrm{N_2O_4(g) + 3CO(g) \longrightarrow N_2O(g) + 3CO_2(g)}$ (in kJ) is (Given : Enthalpies of formation of CO(g), $\mathrm{CO_2(g)}$, $\mathrm{N_2O(g)}$ and $\mathrm{N_2O_4(g)}$ are respectively $-110, -393, 81$ and $9.7\,kJ\,mol^{-1}$) (A) +678 (B) -678 (C) -778 (D) +578

    ›Reveal solutionSolution

    Applying Hess's law ($\Delta_rH = \sum \Delta_fH_{products} - \sum \Delta_fH_{reactants}$) to this reaction gives $\Delta_rH \approx -778$ kJ.

    Concept and Intuition

    By Hess's law, the enthalpy of any reaction can be computed from tabulated standard enthalpies of formation, since enthalpy is a state function:

    $\Delta_rH = \sum n_p\,\Delta_fH^{\ominus}(\text{products}) - \sum n_r\,\Delta_fH^{\ominus}(\text{reactants})$

    Step-by-Step Solution

    1. Reaction: $\mathrm{N_2O_4(g) + 3CO(g) \to N_2O(g) + 3CO_2(g)}$.
    2. Products sum: $\Delta_fH(\mathrm{N_2O}) + 3\,\Delta_fH(\mathrm{CO_2}) = 81 + 3(-393) = 81 - 1179 = -1098$ kJ.
    3. Reactants sum: $\Delta_fH(\mathrm{N_2O_4}) + 3\,\Delta_fH(\mathrm{CO}) = 9.7 + 3(-110) = 9.7 - 330 = -320.3$ kJ. …

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12. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
    Q.The enthalpy of formation of $\mathrm{CO_2(g)}$ and $\mathrm{H_2O}(l)$ are $-393.5$ and $-286$ $\mathrm{kJ\ mol^{-1}}$ respectively. If the heat of combustion of $\mathrm{CH_3OH}(l)$ is $-749\ \mathrm{kJ\ mol^{-1}}$, the enthalpy of formation of $\mathrm{CH_3OH}(l)$ (in $\mathrm{kJ\ mol^{-1}}$) is (A) $+216.5$ (B) $-216.5$ (C) $-355.5$ (D) $+355.5$

    ›Reveal solutionSolution

    This uses Hess's law with the combustion enthalpy to back-calculate the enthalpy of formation of methanol, giving $-216.5\,\mathrm{kJ\,mol^{-1}}$.

    Concept and Intuition

    Hess's law states that $\Delta H_{reaction} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$, regardless of the path. Since $O_2(g)$ is an element in its standard state, its $\Delta H_f = 0$ and drops out.

    Step-by-Step Solution

    1. Combustion reaction: $\mathrm{CH_3OH}(l) + \tfrac{3}{2}\mathrm{O_2}(g) \rightarrow \mathrm{CO_2}(g) + 2\mathrm{H_2O}(l)$.
    2. $\Delta H_{comb} = [\Delta H_f(\mathrm{CO_2}) + 2\Delta H_f(\mathrm{H_2O})] - [\Delta H_f(\mathrm{CH_3OH}) + \tfrac{3}{2}\times0]$.
    3. Substitute: $-749 = [-393.5 + 2(-286)] - \Delta H_f(\mathrm{CH_3OH})$. …

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