Q.If the combustion of 1g of graphite produces 20.7 kJ of heat, what will be molar enthalpy change? Give the significance of sign also.

Showing the 12 most recent of 16 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Consider the following reaction $A_2B_4(g) \rightarrow 2AB_2(g)$ At 300 K, $\Delta_rH$ for this reaction is $-x$ kJ $mol^{-1}$. What is its $\Delta_rU$ (in kJ $mol^{-1}$) at the same temperature? ($R = 8.3\ J\,mol^{-1}\,K^{-1}$) (A) $-(x + 2490)$ (B) $(x + 2490)$ (C) $-(x + 2.49)$ (D) $(x + 2.49)$

   ›Reveal solutionSolution

   This applies the relation between $\Delta H$ and $\Delta U$ for a gas-phase reaction with a change in mole number; $\Delta_rU=-(x+2.49)$ kJ/mol.

   Concept and Intuition

   Enthalpy and internal energy differ by the $pV$-work term. For reactions involving gases, $\Delta H = \Delta U + \Delta n_{gas}\,RT$, where $\Delta n_{gas}$ is the change in moles of gas (products − reactants). This correction matters whenever the number of gas moles changes during the reaction, as it does here (1 mole of $A_2B_4$ becomes 2 moles of $AB_2$).

   Step-by-Step Solution

   1. Reaction: $A_2B_4(g)\rightarrow 2AB_2(g)$. Moles of gas: reactant = 1, product = 2, so $\Delta n_{gas}=2-1=1$.
   2. Relation: $\Delta H = \Delta U + \Delta n_{gas}RT \Rightarrow \Delta U = \Delta H - \Delta n_{gas}RT$. …

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2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.Identify the correct statements from the following I. Bomb calorimeter is used to determine the heat absorbed by water in a chemical reaction at constant volume II. Heat capacity is an extensive property III. The units of entropy are J K The correct answer is (A) I, II, III (B) I, II only (C) I, III only (D) II, III only

   ›Reveal solutionSolution

   Tests three basic thermodynamics facts at once: how a bomb calorimeter works (constant volume), whether heat capacity is extensive or intensive, and the correct SI unit of entropy. Only statements I and II are true.

   Concept and Intuition

   A bomb calorimeter is a strong steel vessel (the "bomb") immersed in a known mass of water, and the whole assembly is sealed — its volume cannot change during the reaction. So any reaction run inside it proceeds at constant volume, and the heat evolved/absorbed measured this way equals $\Delta U$ (internal energy change), not $\Delta H$. The heat released by the reaction flows into the surrounding water (and the calorimeter body), and the resulting temperature rise of the water is what lets us calculate the heat of reaction — so statement I correctly describes the working principle.

   Heat capacity $C = q/\Delta T$ tells you how much heat is needed to raise the temperature of a given sample by 1 K. Double the amount of substance and you double the heat capacity — it scales with the size of the system, which is exactly the definition of an extensive property. (Contrast this with specific heat capacity, $c = C/m$, which is intensive because dividing by mass removes the size-dependence.)

   Entropy $S$ is defined through $dS = \dfrac{dq_{rev}}{T}$ — heat divided by temperature — so its SI unit must be $J\,K^{-1}$ (joule per kelvin), sometimes written per mole as $J\,K^{-1}\,mol^{-1}$. Writing the unit as "$J\,K$" (joule times kelvin) is dimensionally wrong.

   Step-by-Step Solution …

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3. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.At 298 K, the value of $(\Delta H - \Delta U)$ for the combustion of 1 mole $C_4H_{10}(g)$ is $x$ kJ and for the combustion of 1 mole glucose, the value of $(\Delta H - \Delta U)$ is $y$ kJ. The value of $(x-y)$ (in kJ) is ($R = 8.3\ J\ K^{-1}mol^{-1}$) (A) $+8.657$ (B) $-8.657$ (C) $-9.659$ (D) $+4.329$

   ›Reveal solutionSolution

   $(\Delta H-\Delta U)=\Delta n_g RT$; butane's combustion has $\Delta n_g=-3.5$ while glucose's has $\Delta n_g=0$ (glucose is solid), giving $x-y=-8.657$ kJ.

   Concept and Intuition

   $\Delta H$ and $\Delta U$ differ only by the $pV$-work term associated with a change in the number of moles of gas during the reaction: $\Delta H = \Delta U + \Delta n_g RT$, i.e. $\Delta H - \Delta U = \Delta n_g RT$. Only gaseous species count toward $\Delta n_g$; solids and liquids contribute negligible volume change and are ignored.

   Step-by-Step Solution

   1. Butane combustion: $C_4H_{10}(g) + \dfrac{13}{2}O_2(g) \rightarrow 4CO_2(g) + 5H_2O(l)$. Gaseous moles: reactants $=1+6.5=7.5$; products $=4$ (water is liquid). $\Delta n_g = 4-7.5=-3.5$. $x = \Delta n_g RT = -3.5 \times 8.3\times10^{-3}\text{ kJ/Kmol} \times 298\text{ K} = -8.657$ kJ.
   2. Glucose combustion: $C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l)$. …

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4. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.What is the enthalpy change (in $J\ mol^{-1}$) for the conversion of 1 mole of $H_2O\ (l)$ at $10\ °C$ to 1 mole of $H_2O\ (s)$ at $-10\ °C$? (At $0\ °C$, $H_2O\ (s) + x\ kJ\ mol^{-1} \rightarrow H_2O\ (l)$; $C_p(H_2O\ (l)) = y\ J\ mol^{-1}K^{-1}$ $C_p(H_2O\ (s)) = z\ J\ mol^{-1}K^{-1})$ (A) $-(1000x + 10y + 10z)$ (B) $-(x + y + z)$ (C) $-(1000x + y - z)$ (D) $-(1000x - y + z)$

   ›Reveal solutionSolution

   Break the process into three Hess's-law steps (cool liquid, freeze, cool solid) and add the enthalpies; converting $x$ from kJ to J gives $1000x$.

   Concept and Intuition

   Enthalpy is a state function, so we can go from $H_2O(l, 10°C)$ to $H_2O(s, -10°C)$ via any convenient path and add up $\Delta H$ for each leg (Hess's law). The natural path passes through $0°C$, the melting/freezing point, where the phase-change enthalpy is defined.

   Step-by-Step Solution

   1. Step 1 — cool the liquid from $10°C$ to $0°C$: $\Delta H_1 = C_p(l)\times \Delta T = y \times (0-10) = -10y \text{ J/mol}$.
   2. Step 2 — freeze at $0°C$: Given $H_2O(s) + x\ \text{kJ/mol} \rightarrow H_2O(l)$ (melting absorbs $x$ kJ/mol), the reverse (freezing) releases the same amount: $\Delta H_2 = -x\ \text{kJ/mol} = -1000x\ \text{J/mol}$.
   3. Step 3 — cool the solid from $0°C$ to $-10°C$: $\Delta H_3 = C_p(s)\times \Delta T = z\times(-10-0) = -10z\ \text{J/mol}$. …

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5. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.The energy required to increase the temperature of 180 g of liquid water from $10\,°C$ to $15\,°C$ is 3765 J. What is $C_p$ of water in $\mathrm{J\, mol^{-1}\, K^{-1}}$? ($\mathrm{H_2O} = 18$u) (A) 75.3 (B) 376.5 (C) 753 (D) 37.65

   ›Reveal solutionSolution

   This is a direct application of $q = nC_p\Delta T$ to find the per-mole heat capacity of liquid water; converting grams to moles and dividing gives $C_p = 75.3\ \text{J mol}^{-1}\text{K}^{-1}$.

   Concept and Intuition

   Heat capacity relates the heat supplied to the resulting temperature change: $q = nC_p\Delta T$ at constant pressure (heating open to atmosphere, as is implicit for a liquid being warmed). Since $C_p$ is asked in J per MOLE per K, the mass of water must first be converted to moles using its molar mass.

   Step-by-Step Solution

   1. Convert mass to moles: $n = \dfrac{180\ \text{g}}{18\ \text{g/mol}} = 10$ mol.
   2. Compute the temperature change: $\Delta T = 15°C - 10°C = 5\ \text{K}$ (a temperature difference in °C equals the same difference in K).
   3. Apply $q = nC_p\Delta T \Rightarrow C_p = \dfrac{q}{n\Delta T} = \dfrac{3765\ \text{J}}{10\ \text{mol}\times 5\ \text{K}} = \dfrac{3765}{50}\ \text{J mol}^{-1}\text{K}^{-1} = 75.3\ \text{J mol}^{-1}\text{K}^{-1}$.
   4. This matches option (A) directly. …

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6. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.For which reaction is $\Delta H \neq \Delta U$ ? (g = gas) (A) $\mathrm{H_2(g) + I_2(g) \longrightarrow 2HI(g)}$ (B) $\mathrm{2NO(g) \longrightarrow N_2(g) + O_2(g)}$ (C) $\mathrm{N_2(g) + 3H_2(g) \longrightarrow 2NH_3(g)}$ (D) $\mathrm{C(s) + O_2(g) \longrightarrow CO_2(g)}$

   ›Reveal solutionSolution

   This tests when $\Delta H \neq \Delta U$ for a reaction — only when the number of gas moles changes; the answer is N₂ + 3H₂ → 2NH₃, option (C).

   Concept and Intuition

   The relation between enthalpy and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ is the change in the number of MOLES OF GAS between products and reactants. If $\Delta n_g = 0$, then $\Delta H = \Delta U$ exactly; only reactions where gas moles change on both sides give $\Delta H \neq \Delta U$.

   Step-by-Step Solution

   1. (A) $\mathrm{H_2(g)+I_2(g)\to 2HI(g)}$: gas moles $2 \to 2$, $\Delta n_g = 0$ → $\Delta H = \Delta U$.
   2. (B) $\mathrm{2NO(g)\to N_2(g)+O_2(g)}$: gas moles $2 \to 2$, $\Delta n_g=0$ → $\Delta H=\Delta U$.
   3. (C) $\mathrm{N_2(g)+3H_2(g)\to 2NH_3(g)}$: gas moles $4 \to 2$, $\Delta n_g = 2-4 = -2 \neq 0$ → $\Delta H \neq \Delta U$. …

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7. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
   Q.Given below are two statements Statement-I: $\Delta U$ can be measured by bomb calorimeter Statement-II: Heat is not transferred from calorimeter to surroundings The correct answer is (A) Both Statement-I and Statement-II are correct (B) Both Statement-I and Statement-II are not correct (C) Statement-I is correct but Statement-II is not correct (D) Statement-I is not correct but Statement-II is correct

   ›Reveal solutionSolution

   Bomb calorimetry directly measures $\Delta U$ (constant-volume heat), not $\Delta H$, and the calorimeter is designed to be adiabatic so that no heat is exchanged with the surroundings. Both statements are correct.

   Concept & Intuition: Bomb Calorimetry and Enthalpy

   A bomb calorimeter is a sealed, rigid container (the "bomb") placed inside a water bath, all insulated from the outside. Because the bomb is rigid, the volume of the reaction mixture cannot change — so no pressure-volume work is done. The first law of thermodynamics tells us that for a constant-volume process, the heat absorbed or released equals the change in internal energy: $q_V = \Delta U$. That's exactly what the calorimeter measures: the temperature change of the water bath tells us $\Delta U$ for the reaction.

   Enthalpy $\Delta H$ is defined as $\Delta H = \Delta U + \Delta (PV)$. For reactions involving gases, $\Delta (PV)$ can be significant, so $\Delta H$ and $\Delta U$ differ. Bomb calorimetry gives $\Delta U$ directly; $\Delta H$ is then calculated by adding $P\Delta V$ (or $\Delta n_g RT$ for ideal gases). The second statement is about the calorimeter's design: it is adiabatic — meaning it prevents heat transfer to or from the surroundings — so all the heat from the reaction stays inside the calorimeter, allowing accurate measurement.

   Now, let's check each statement step by step.

   1. Statement-I: "$\Delta U$ can be measured by bomb calorimeter"

      In a bomb calorimeter, the reaction occurs at constant volume. Under constant volume, $W = 0$ (no expansion work), so from the first law $\Delta U = Q + W$, we get $\Delta U = Q_V$. The calorimeter measures $Q_V$ by the temperature change of the water. Therefore, $\Delta U$ is directly obtained.

      Conclusion: Statement-I is correct.

   2. Statement-II: "Heat is not transferred from calorimeter to surroundings" …

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8. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.Observe the following reaction $\mathrm{ABO_3(s) \xrightarrow{1000\,K} AO(s) + BO_2(g)}$ $\Delta_r H$ for this reaction is $x\ \mathrm{kJ\,mol^{-1}}$. What is its $\Delta_r U$ (in $\mathrm{kJ\,mol^{-1}}$) at the same temperature? ($R = 8.3\ \mathrm{J\,mol^{-1}\,K^{-1}}$) (A) $x - 8300$ (B) $x + 8.3$ (C) $x + 8300$ (D) $x - 8.3$

   ›Reveal solutionSolution

   With $\Delta n_g = 1$ at 1000 K, $\Delta_r U = \Delta_r H - \Delta n_g RT = x - 8.3\ \mathrm{kJ/mol}$.

   Concept and Intuition

   Enthalpy and internal energy of reaction are related by $\Delta H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ is the change in moles of gaseous species only (solids/liquids don't contribute to the $PV$ work term for an ideal-gas approximation).

   Step-by-Step Solution

   1. Identify gaseous moles: reactant $\mathrm{ABO_3(s)}$ is solid (0 mol gas); products are $\mathrm{AO(s)}$ (solid, 0 mol gas) and $\mathrm{BO_2(g)}$ (1 mol gas). So $\Delta n_g = 1 - 0 = 1$.
   2. Compute $\Delta n_g RT$ at $T=1000\ \mathrm{K}$: $\Delta n_g RT = 1 \times 8.3\ \mathrm{J\,mol^{-1}K^{-1}} \times 1000\ \mathrm{K} = 8300\ \mathrm{J/mol} = 8.3\ \mathrm{kJ/mol}$. …

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9. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.At 300 K the enthalpy change for the following reaction is $-2800\ \mathrm{kJ\ mol^{-1}}$ $\mathrm{C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l)}$ What is the $\Delta U$ for the same reaction at 300 K ($\mathrm{kJ\ mol^{-1}}$)? (A) -2802.49 (B) -2800.00 (C) -2814.94 (D) +2802.49

   ›Reveal solutionSolution

   Because the moles of gas are unchanged ($\Delta n_g=0$) in this combustion, $\Delta U$ equals $\Delta H$ exactly.

   Concept and Intuition

   $\Delta H=\Delta U+\Delta n_g\,RT$, where $\Delta n_g$ counts only gaseous species (moles of gaseous products minus moles of gaseous reactants). Solids and liquids don't contribute to $\Delta n_g$.

   Step-by-Step Solution

   1. Reaction: $C_6H_{12}O_6(s)+6O_2(g)\rightarrow 6CO_2(g)+6H_2O(l)$.
   2. Gaseous reactants: $6\,O_2$. Gaseous products: $6\,CO_2$. ($C_6H_{12}O_6$ is solid, $H_2O$ is liquid — neither counts.)
   3. $\Delta n_g=6-6=0$.
   4. $\Delta H=\Delta U+\Delta n_g RT=\Delta U+0=\Delta U$.
   5. So $\Delta U=\Delta H=-2800.00\,\text{kJ/mol}$. …

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10. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
    Q.When 'X' g of graphite is completely burnt in a bomb calorimeter in excess of $\mathrm{O_2}$ at 298 K and 1 atm pressure as given in the equation $\mathrm{C(graphite) + O_2(g) \to CO_2(g)}$. The temperature of calorimeter raised from 298 K to 302 K. If the heat capacity of the calorimeter and molar enthalpy change for the reaction at 1 atm and 298 K are 20.7 kJ K$^{-1}$ and -248.4 kJ mol$^{-1}$, 'X' in g is (A) 8 (B) 2 (C) 3 (D) 4

    ›Reveal solutionSolution

    A bomb calorimeter measures heat at constant volume ($q_V=\Delta U$); since this reaction has no change in gaseous moles, $\Delta U=\Delta H$, letting us convert the temperature rise directly into moles of graphite burnt.

    Concept and Intuition

    A bomb calorimeter is a rigid, sealed vessel — the reaction occurs at constant volume, so the heat measured is $q_V=\Delta U$, not $\Delta H$ (which is the constant-pressure heat, $q_P$). In general $\Delta H=\Delta U+\Delta n_{gas}RT$. Here $\mathrm{C(graphite)+O_2(g)\to CO_2(g)}$ has one mole of gas ($O_2$) turning into one mole of gas ($CO_2$), so $\Delta n_{gas}=0$ and $\Delta H=\Delta U$ exactly. That means the given molar enthalpy change (−248.4 kJ/mol) is also the heat released per mole at constant volume, and can be used directly with the calorimeter data.

    Step-by-Step Solution

    1. Heat absorbed by the calorimeter: $q=C_{cal}\times\Delta T = 20.7\ \text{kJ K}^{-1}\times(302-298)\text{K} = 20.7\times4 = 82.8\ \text{kJ}$.
    2. This heat came from the combustion of graphite, whose magnitude per mole is $248.4$ kJ (since $\Delta U=\Delta H$ here). …

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11. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
    Q.If standard molar enthalpy change and standard molar internal energy change measured in bomb calorimeter are equal, which one of the following statements is correct? (A) $\Delta n > 0$, with increase in pressure (B) $\Delta n > 0$, with decrease in pressure (C) $\Delta n < 0$, with increase in pressure (D) $\Delta n = 0$, at constant pressure

    ›Reveal solutionSolution

    $\Delta H=\Delta U$ only when the reaction involves no net change in gaseous moles, $\Delta n_g=0$.

    Concept and Intuition

    The relation between enthalpy change and internal energy change for a reaction involving gases is $\Delta H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ is the change in the number of moles of gaseous species (products $-$ reactants). $\Delta H$ is naturally the quantity measured at constant pressure, while a bomb calorimeter (constant volume) measures $\Delta U$ directly. The two coincide only when the correction term $\Delta n_g RT$ vanishes, i.e. $\Delta n_g = 0$.

    Step-by-Step Solution

    1. Write the general relation: $\Delta H = \Delta U + \Delta n_g RT$.
    2. For $\Delta H = \Delta U$, we need $\Delta n_g RT = 0$.
    3. Since $R$ and $T$ are non-zero, this requires $\Delta n_g = 0$. …

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12. AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
    Q.Combustion of 1 mole graphite releases $+2.48 \times 10^{2}$ kJ of energy. What will be the temperature of a bomb calorimeter, if 1 g of graphite is burnt at 298 K. The heat capacity of the bomb calorimeter is 10.35 kJ/K (A) 298 K (B) 296 K (C) 300 K (D) 299 K

    ›Reveal solutionSolution

    Scaling the molar combustion energy down to 1 g of graphite and dividing by the calorimeter's heat capacity gives a temperature rise of about 2 K, so the final temperature is 300 K.

    Concept and Intuition

    In bomb calorimetry, the heat released by a combustion reaction is absorbed entirely by the (constant-volume) calorimeter assembly, raising its temperature according to $q = C\Delta T$, where $C$ is the calorimeter's total heat capacity. Since the given combustion enthalpy is per mole, we must first convert it to the heat released by the actual mass burnt (1 g here).

    Step-by-Step Solution

    1. Heat released per mole of graphite: $2.48\times10^2\ \mathrm{kJ} = 248\ \mathrm{kJ/mol}$.
    2. Molar mass of graphite (carbon) $=12\ \mathrm{g/mol}$, so heat released per gram: $248/12 = 20.667\ \mathrm{kJ}$. …

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