Q.Calculate $\Delta_r G^\ominus$ for the conversion of oxygen to ozone, $\tfrac{3}{2} O_2(g) \rightarrow O_3(g)$ at 298 K, if $K_p$ for this conversion is $2.47 \times 10^{-29}$.

Showing the 12 most recent of 17 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.The $\Delta_f H^\Theta$ of BaCO$_3$ (s), BaO (s) and CO$_2$

   (g) is respectively $-1216.3$, $-553.5$ and $-393.5$ kJ mol$^{-1}$. What is the value of $x$ (in kJ mol$^{-1}$) in the following reaction? BaCO$_3$ (s) $\xrightarrow{\Delta}$ BaO (s) + CO$_2$

   (g) $- x$ (A) $-269.3$ (B) $269.3$ (C) $2163$ (D) $-2163$

   ›Reveal solutionSolution

   $\Delta_r H^\Theta = \Delta_f H(BaO) + \Delta_f H(CO_2) - \Delta_f H(BaCO_3) = +269.3\text{ kJ mol}^{-1}$, so $x = 269.3$.

   For $\text{BaCO}_3(s)\rightarrow \text{BaO}(s) + \text{CO}_2(g)$:

   $\Delta_r H^\Theta = \big[\Delta_f H(\text{BaO}) + \Delta_f H(\text{CO}_2)\big] - \Delta_f H(\text{BaCO}_3)$

   $\Delta_r H^\Theta = \big[(-553.5) + (-393.5)\big] - (-1216.3) = -947.0 + 1216.3 = +269.3\ \text{kJ mol}^{-1}$

   The decomposition is endothermic; the energy term written as $-x$ in the equation corresponds to $x = +269.3\ \text{kJ mol}^{-1}$.

   ✓Final answer

   $x = 269.3\ \text{kJ mol}^{-1}$ — option (B).

2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.The transition temperature of sulphur is (A) $369~^\circ \mathrm{C}$ (B) $369~\mathrm{K}$ (C) $133~\mathrm{K}$ (D) $133~^\circ \mathrm{C}$

   ›Reveal solutionSolution

   The transition temperature of sulphur (rhombic to monoclinic) is a well‑known constant of 369 K (≈ 96 °C). The correct option is (B).

   The question asks for the transition temperature of sulphur. This refers to the temperature at which the solid phase changes from the rhombic (α‑sulphur) to the monoclinic (β‑sulphur) form. This is a classic fact from solid‑state chemistry and phase diagrams. The key is to recall the value and its correct unit — many students confuse the Celsius and Kelvin scales.

   1. Recall the known value

      The rhombic‑to‑monoclinic transition for sulphur occurs at 96 °C under standard pressure. This is a standard datum in textbooks.

   2. Convert to Kelvin

      Since Kelvin = Celsius + 273.15, we have:

   $96 + 273.15 \approx 369\ \text{K}$

   (Often rounded to 369 K.)

   3. Match with the options
      • (A) 369 °C → far too high (sulphur melts at ~115 °C).
      • (B) 369 K → matches the conversion.
      • (C) 133 K → that’s about –140 °C, far too low.
      • (D) 133 °C → above the melting point, not a solid‑solid transition.

   Watch out

   A common mistake is to pick 369 °C because the number “369” looks familiar, but that temperature would be well above sulphur’s boiling point. Always check the unit.

   Tip

   Remember the mnemonic: “Sulphur switches at 96 °C, which is 369 K” — the digits 3‑6‑9 are easy to recall.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

3. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.The difference between energy of an activated complex and the average energy of reactants is called? (A) Threshold energy (B) Lattice energy (C) Activation energy (D) Kinetic energy

   ›Reveal solutionSolution

   The question asks for the term that describes the energy difference between the activated complex (transition state) and the average energy of reactants. That term is activation energy, so the correct option is (C).

   The key concept here is reaction energy profiles and the transition state theory. In a chemical reaction, reactants must overcome an energy barrier to form products. The highest-energy, unstable arrangement along the reaction path is called the activated complex (or transition state). The energy required to go from the average energy of the reactants up to that peak is precisely the activation energy — it’s the “hump” that must be surmounted.

   Let’s break down why each option fits or doesn’t:

   1. Understanding the definition

      The difference between the energy of the activated complex and the average energy of the reactants is, by definition, the minimum energy that reactants must acquire to react. This is the activation energy ($E_a$). It is always positive for an endothermic or exothermic reaction (except in barrierless reactions, where it’s zero).

   2. Why not (A) Threshold energy?

      Threshold energy is a related but distinct concept from collision theory: it is the minimum kinetic energy that colliding molecules must have for a reaction to occur. While it is numerically equal to activation energy in some simple models, the question specifically asks for the difference between the energy of the activated complex and the average energy of reactants — that is the textbook definition of activation energy, not threshold energy. Threshold energy is defined relative to the ground state of reactants, not their average energy.

   3. Why not (B) Lattice energy?

      Lattice energy is the energy released when gaseous ions form a solid ionic lattice (or the energy required to break it). It has nothing to do with reaction barriers or activated complexes.

   4. Why not (D) Kinetic energy?

      Kinetic energy is the energy of motion. While reactants need kinetic energy to overcome the activation barrier, the difference in potential energies (activated complex minus reactants) is a potential energy difference, not kinetic energy.

   Tip

   A common pitfall is confusing “threshold energy” with “activation energy.” Remember: threshold energy is a minimum kinetic energy for collision, while activation energy is a potential energy difference on the reaction coordinate diagram. The question’s phrasing (“difference between energy of activated complex and average energy of reactants”) points directly to activation energy.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

4. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.What is the enthalpy change (in kJ mol$^{-1}$) for the following reaction? CCl$_4$(g) $\rightarrow$ C(g) + 4Cl(g) ($\Delta_{\text{vap}}$H$^\Theta$(CCl$_4$) = 30.5 kJ mol$^{-1}$ ; $\Delta_{\text{f}}$H$^\Theta$(CCl$_4$) = $-$135.5 kJ mol$^{-1}$ ; $\Delta_{\text{a}}$H$^\Theta$(C) = 715.0 kJ mol$^{-1}$ ; $\Delta_{\text{a}}$H$^\Theta$(Cl$_2$) = 242 kJ mol$^{-1}$) (A) 1304 (B) 326 (C) 1033 (D) 1199

   ›Reveal solutionSolution

   The enthalpy change for atomising CCl$_4$(g) into gaseous atoms is found by reversing its formation from atoms and adding the atomisation enthalpies of C and Cl$_2$. The result is 1304 kJ mol$^{-1}$, option (A).

   The reaction CCl$_4$(g) $\rightarrow$ C(g) + 4Cl(g) is the complete atomisation of gaseous carbon tetrachloride. We are not measuring this directly — we build it from known thermodynamic data using Hess’s law.

   The key idea: the standard enthalpy of formation $\Delta_f H^\Theta$ of a compound is the enthalpy change when it is made from its elements in their standard states. For CCl$_4$(l), the standard state of carbon is graphite (C(s)) and of chlorine is Cl$_2$(g). But our target starts from CCl$_4$(g) and ends with gaseous atoms. So we need to:

   • start with CCl$_4$(g),
   • break it down into its elements in standard states (reverse of formation),
   • then atomise those elements into gaseous atoms.

   We are given $\Delta_{\text{vap}}H^\Theta$ for CCl$_4$ — that tells us the liquid-to-gas step, because the formation enthalpy given is for the liquid, not the gas. Let’s work through it.

   1. Convert liquid CCl$_4$ to gaseous CCl$_4$

      The given $\Delta_f H^\Theta(\text{CCl}_4) = -135.5$ kJ mol$^{-1}$ is for the liquid. Our target reaction starts with the gas, so we must first vaporise it:

      $\text{CCl}_4(l) \rightarrow \text{CCl}_4(g)$, $\Delta H = +30.5$ kJ mol$^{-1}$ (endothermic).

   2. Reverse the formation of CCl$_4$(g) from its elements

      The formation reaction is:

      $\text{C(s)} + 2\text{Cl}_2(g) \rightarrow \text{CCl}_4(l)$, $\Delta H = -135.5$ kJ mol$^{-1}$.

      But we need the gas, so combine step 1:

      $\text{C(s)} + 2\text{Cl}_2(g) \rightarrow \text{CCl}_4(g)$, $\Delta H = -135.5 + 30.5 = -105.0$ kJ mol$^{-1}$.

      Reversing this gives:

      $\text{CCl}_4(g) \rightarrow \text{C(s)} + 2\text{Cl}_2(g)$, $\Delta H = +105.0$ kJ mol$^{-1}$.

   3. Atomise carbon

      $\text{C(s)} \rightarrow \text{C(g)}$, $\Delta_{\text{a}}H^\Theta(\text{C}) = 715.0$ kJ mol$^{-1}$.

   4. Atomise chlorine

      $\text{Cl}_2(g) \rightarrow 2\text{Cl}(g)$, $\Delta_{\text{a}}H^\Theta(\text{Cl}_2) = 242$ kJ mol$^{-1}$.

      We have 2 moles of Cl$_2$ from step 2, so this contributes $2 \times 242 = 484$ kJ mol$^{-1}$.

   5. Add all steps

      Total enthalpy change = $+105.0 + 715.0 + 484 = 1304$ kJ mol$^{-1}$.

   Watch out

   A common mistake is to forget that $\Delta_f H^\Theta(\text{CCl}_4)$ is for the liquid, not the gas. Using $-135.5$ directly without adding the vaporisation enthalpy gives $135.5 + 715 + 484 = 1334.5$, which is not among the options — but close enough to trap you. Always check the physical state.

   Tip

   You can think of this as: enthalpy of atomisation of CCl$_4$(g) = $-\Delta_f H^\Theta(\text{CCl}_4,g) + \Delta_{\text{a}}H^\Theta(\text{C}) + 2 \times \Delta_{\text{a}}H^\Theta(\text{Cl}_2)$. Here $\Delta_f H^\Theta(\text{CCl}_4,g) = -135.5 + 30.5 = -105$, so the answer is $105 + 715 + 484 = 1304$.

   ✓Final answer

   The enthalpy change is $\boxed{1304\ \text{kJ mol}^{-1}}$, which corresponds to option (A).

5. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.When electromagnetic radiation of wavelength 310 nm falls on the surface of a metal having work function 3.55 eV, the velocity of photoelectrons emitted is $x \times 10^5 \ \mathrm{ms}^{-1}$. The value of $x$ is (Nearest integer) ($m_e = 9 \times 10^{-31} \ \mathrm{kg}$) (A) 2 (B) 4 (C) 5 (D) 6

   ›Reveal solutionSolution

   The photoelectric effect gives the kinetic energy of emitted electrons as the photon energy minus the work function. Using the given wavelength and work function, the electron velocity is found to be about $4 \times 10^5$ m/s, so the nearest integer $x$ is 4.

   Concept & Intuition

   This is a classic photoelectric effect problem. When light of a certain frequency (or wavelength) hits a metal, each photon can transfer its energy to an electron. If the photon’s energy exceeds the metal’s work function (the minimum energy needed to free an electron), the excess becomes the electron’s kinetic energy. The kinetic energy then determines the electron’s speed via $K = \frac12 m v^2$. The key is to convert all units consistently — here, photon energy is in joules (from wavelength) and work function is given in electronvolts, so we must convert eV to joules.

   Step-by-step solution

   1. Find the photon energy in joules The energy of a photon is $E = \frac{hc}{\lambda}$, where $h = 6.63 \times 10^{-34} \ \text{J·s}$, $c = 3 \times 10^8 \ \text{m/s}$, $\lambda = 310 \ \text{nm} = 310 \times 10^{-9} \ \text{m}$.

   $E = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{310 \times 10^{-9}} = \frac{1.989 \times 10^{-25}}{3.1 \times 10^{-7}} \approx 6.416 \times 10^{-19} \ \text{J}.$

   2. Convert the work function to joules Work function $\phi = 3.55 \ \text{eV}$. Since $1 \ \text{eV} = 1.6 \times 10^{-19} \ \text{J}$,

   $\phi = 3.55 \times 1.6 \times 10^{-19} = 5.68 \times 10^{-19} \ \text{J}.$

   3. Compute the kinetic energy of the photoelectron By Einstein’s photoelectric equation:

   $K = E - \phi = (6.416 - 5.68) \times 10^{-19} = 0.736 \times 10^{-19} \ \text{J}.$

   4. Relate kinetic energy to velocity $K = \frac12 m_e v^2$, with $m_e = 9 \times 10^{-31} \ \text{kg}$.

   $v^2 = \frac{2K}{m_e} = \frac{2 \times 0.736 \times 10^{-19}}{9 \times 10^{-31}} = \frac{1.472 \times 10^{-19}}{9 \times 10^{-31}} = 1.6356 \times 10^{11}.$

   Taking the square root:

   $v = \sqrt{1.6356 \times 10^{11}} \approx 4.044 \times 10^5 \ \text{m/s}.$

   5. Identify $x$ The velocity is given as $x \times 10^5 \ \text{m/s}$, so $x \approx 4.044$. The nearest integer is $4$.

   Watch out

   A common mistake is forgetting to convert the work function from eV to joules, or using the wavelength in nm without converting to meters. Always check units before calculating.

   Tip

   You can also work entirely in eV and then convert kinetic energy to joules at the end: $E_{\text{photon}} = \frac{1240}{\lambda (\text{nm})} \ \text{eV} \approx \frac{1240}{310} = 4.0 \ \text{eV}$, so $K = 4.0 - 3.55 = 0.45 \ \text{eV} = 0.45 \times 1.6 \times 10^{-19} \ \text{J}$, then proceed as above — same result.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

6. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.At 298 K the equilibrium constant for the reaction M(s) + 2Ag$^+$(aq) $\rightarrow$ M$^{2+}$(aq) + 2Ag(s) is $10^{15}$. What is the $E^\Theta_{cell}$ (in V) for this reaction? $\left(\dfrac{2.303RT}{F}\right) = 0.06$ V (A) 0.45 (B) 0.90 (C) 0.225 (D) 1.10

   ›Reveal solutionSolution

   The key idea is to relate the equilibrium constant to the standard cell potential using the Nernst equation at equilibrium. The result is $E^\Theta_{\text{cell}} = 0.45\ \text{V}$, which corresponds to option (A).

   We are given a reaction between a metal M and silver ions, with an enormous equilibrium constant ($10^{15}$). That tells us the reaction strongly favors products — so the standard cell potential must be positive and fairly large. The bridge between $K$ and $E^\Theta$ is the Nernst equation at equilibrium, where the cell potential is zero and the reaction quotient equals $K$.

   1. Recall the fundamental relation At equilibrium, the Nernst equation becomes:

   $E^\Theta_{\text{cell}} = \frac{0.0591}{n} \log K \quad (\text{at 298 K, using } \frac{2.303RT}{F} = 0.0591\ \text{V})$

   But here they give $\frac{2.303RT}{F} = 0.06\ \text{V}$ (a rounded value), so we use:

   $E^\Theta_{\text{cell}} = \frac{0.06}{n} \log K$

   where $n$ is the number of moles of electrons transferred in the balanced reaction.

   2. Determine $n$ The reaction is:

   $\text{M(s)} + 2\text{Ag}^+(aq) \rightarrow \text{M}^{2+}(aq) + 2\text{Ag(s)}$

   Each Ag⁺ gains one electron to become Ag, and two Ag⁺ ions are reduced, so total electrons transferred = 2. Also, M loses two electrons to become M²⁺. Hence $n = 2$.

   3. Plug in the numbers

   $E^\Theta_{\text{cell}} = \frac{0.06}{2} \log(10^{15}) = 0.03 \times 15 = 0.45\ \text{V}$

   4. Check the options 0.45 V matches option (A).

   Tip

   A common mistake is to forget that $n$ is the total number of electrons transferred, not the coefficient of a species. Here, even though the coefficient of Ag⁺ is 2, the electrons transferred per formula unit is also 2 — so it’s consistent.

   Watch out

   If you mistakenly used $n = 1$ (thinking of one electron per Ag⁺ but forgetting there are two Ag⁺), you’d get $0.06 \times 15 = 0.90\ \text{V}$, which is option (B) — a tempting distractor.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

7. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.At T(K) 10 L of an ideal gas was expanded to 12 L against a pressure of 2 atm irreversibly. What is the work done by the gas? (A) $-4$ atm L (B) Zero (C) $-240$ atm L (D) $-4$ J

   ›Reveal solutionSolution

   For an irreversible expansion against a constant external pressure, the work done by the gas is calculated using the external pressure and the change in volume. The work done is $\boxed{-4 \text{ atm L}}$.

   When a gas expands or contracts, it does work on its surroundings or has work done on it by the surroundings. The amount of work done depends on the path taken by the process.

   For an irreversible process, especially when the expansion or compression occurs against a constant external pressure, the work done is determined by this external pressure, not the internal pressure of the gas. This is because the system (the gas) is not in equilibrium with its surroundings throughout the process. The external pressure is the opposing force that the gas must overcome to expand.

   The work done ($W$) by a gas during an irreversible expansion or compression against a constant external pressure ($P_{ext}$) is given by:

   $W = -P_{ext} \Delta V$

   where $\Delta V = V_2 - V_1$ is the change in volume.

   The negative sign indicates that if the gas expands ($\Delta V > 0$), it does work on the surroundings, and $W$ is negative. If the gas is compressed ($\Delta V < 0$), work is done on the gas by the surroundings, and $W$ is positive.

   Here, the gas is expanding, so we expect the work done by the gas to be negative.

   1. Identify the given values:

      • Initial volume, $V_1 = 10 \text{ L}$
      • Final volume, $V_2 = 12 \text{ L}$
      • External pressure, $P_{ext} = 2 \text{ atm}$
      • The process is irreversible.

   2. Calculate the change in volume ($\Delta V$):

      The change in volume is the final volume minus the initial volume.

      $\Delta V = V_2 - V_1 = 12 \text{ L} - 10 \text{ L} = 2 \text{ L}$

   3. Apply the formula for work done in an irreversible process:

      Using the formula $W = -P_{ext} \Delta V$:

      $W = -(2 \text{ atm}) \times (2 \text{ L})$

      $W = -4 \text{ atm L}$

      The unit "atm L" (atmosphere-litre) is a valid unit of energy, commonly used in thermodynamics problems involving pressure and volume.

   Watch out

   Do not confuse irreversible work with reversible work. For a reversible process, the work done is calculated using the internal pressure of the gas, which changes continuously, typically requiring integration ($W = -\int P_{internal} dV$). For an irreversible process against a constant external pressure, only the external pressure matters.

   The calculated work done is $-4 \text{ atm L}$. Comparing this with the given options, option (A) matches our result.

   ✓Final answer

   The work done by the gas is $\boxed{-4 \text{ atm L}}$.

8. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.The equilibrium constant for the reaction M(s) + 2Ag$^+$(aq) $\rightarrow$ M$^{2+}$(aq) + 2Ag(s) is $10^{15}$. What is the Gibbs energy change ($\Delta_r G^\Theta$ in KJ mol$^{-1}$) for this reaction? $\left(\dfrac{RT}{F} \times 2.303 = 0.06\,\text{V};\, F = 96500\,\text{C mol}^{-1}\right)$. (A) $-86850$ (B) $-96500$ (C) $-86.85$ (D) $-96.5$

   ›Reveal solutionSolution

   The key is linking the equilibrium constant to the standard Gibbs energy via $\Delta_r G^\Theta = -RT \ln K$, and then using the given relation $RT/F \times 2.303 = 0.06$ V to compute the numerical value. The answer is -86.85 kJ mol⁻¹, which is option (C).

   The problem gives you the equilibrium constant $K = 10^{15}$ for a redox reaction and asks for $\Delta_r G^\Theta$ in kJ mol⁻¹. The direct thermodynamic link is the equation $\Delta_r G^\Theta = -RT \ln K$. But here they’ve supplied a specific value for $\frac{RT}{F} \times 2.303 = 0.06$ V, which is a hint that you can avoid plugging in $R$ and $T$ separately — just use this conversion factor.

   Notice that $\ln K = 2.303 \log_{10} K$, so $\Delta_r G^\Theta = -RT \times 2.303 \log_{10} K$. And since $\frac{RT}{F} \times 2.303 = 0.06$ V, you have $RT \times 2.303 = 0.06 F$. This lets you compute $\Delta_r G^\Theta$ directly in joules, then convert to kJ.

   1. Write the standard relation:

      $\Delta_r G^\Theta = -RT \ln K = -RT \times 2.303 \log_{10} K$.

   2. Substitute $\log_{10} K = \log_{10} (10^{15}) = 15$:

      $\Delta_r G^\Theta = -RT \times 2.303 \times 15$.

   3. Use the given: $RT \times 2.303 = 0.06 \times F$, where $F = 96500$ C mol⁻¹. So:

      $\Delta_r G^\Theta = - (0.06 \times 96500) \times 15$ J mol⁻¹.

   4. Compute stepwise:

      $0.06 \times 96500 = 5790$ J mol⁻¹.

      Then $5790 \times 15 = 86850$ J mol⁻¹.

      So $\Delta_r G^\Theta = -86850$ J mol⁻¹.

   5. Convert to kJ mol⁻¹: divide by 1000 → $-86.85$ kJ mol⁻¹.

   Watch out

   A common mistake is to forget the conversion from J to kJ and pick option (A) $-86850$, which is the value in joules, not kilojoules. Always check the units asked for in the question.

   Tip

   The given factor $\frac{RT}{F} \times 2.303 = 0.06$ V is actually the value of $\frac{2.303RT}{F}$ at around 298 K — it’s the same constant used in the Nernst equation. Recognizing this saves you from needing $R$ and $T$ separately.

   ✓Final answer

   The Gibbs energy change is $\boxed{-86.85\ \text{kJ mol}^{-1}}$, which corresponds to option (C).

9. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.The standard enthalpy of combustion of C (graphite), H$_2$

   (g) and CH$_3$OH(l) respectively are $-393$, $-286$ and $-726$ kJ mol$^{-1}$. What is the standard enthalpy of formation of methanol? (A) $-726$ kJ mol$^{-1}$ (B) $-239$ kJ mol$^{-1}$ (C) $-96$ kJ mol$^{-1}$ (D) $+96$ kJ mol$^{-1}$

   ›Reveal solutionSolution

   Use Hess’s law: the enthalpy of formation of methanol is found by combining the combustion enthalpies of its elements and the combustion enthalpy of methanol. The result is $-239\ \text{kJ mol}^{-1}$, option (B).

   The key idea here is Hess’s law: enthalpy change for a reaction is the same whether it happens directly or in steps. We want the enthalpy of formation of methanol, i.e., the enthalpy change for:

   $\text{C(graphite)} + 2\,\text{H}_2(g) + \tfrac12\,\text{O}_2(g) \rightarrow \text{CH}_3\text{OH}(l)$

   We are given combustion enthalpies, which are enthalpies of complete burning in oxygen. So we can think of building the formation reaction from combustion steps.

   ---

   1. Write the target formation reaction

   $\text{C(s)} + 2\,\text{H}_2(g) + \tfrac12\,\text{O}_2(g) \rightarrow \text{CH}_3\text{OH}(l) \quad \Delta H_f^\circ = ?$

   2. Write the given combustion reactions
      • Combustion of C(graphite):

   $\text{C(s)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H = -393\ \text{kJ/mol}$

   • Combustion of H₂:

   $\text{H}_2(g) + \tfrac12\,\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H = -286\ \text{kJ/mol}$

   • Combustion of methanol:

   $\text{CH}_3\text{OH}(l) + \tfrac32\,\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\,\text{H}_2\text{O}(l) \quad \Delta H = -726\ \text{kJ/mol}$

   3. Reverse the methanol combustion — because we want methanol as a product, not a reactant. Reversing changes the sign:

   $\text{CO}_2(g) + 2\,\text{H}_2\text{O}(l) \rightarrow \text{CH}_3\text{OH}(l) + \tfrac32\,\text{O}_2(g) \quad \Delta H = +726\ \text{kJ/mol}$

   4. Add the combustion of C and 2× combustion of H₂ to the reversed methanol combustion, so that CO₂ and H₂O cancel.
      • C combustion:

   $\text{C(s)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H = -393$

   • 2 × H₂ combustion:

   $2\,\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\,\text{H}_2\text{O}(l) \quad \Delta H = 2 \times (-286) = -572$

   • Reversed methanol combustion:

   $\text{CO}_2(g) + 2\,\text{H}_2\text{O}(l) \rightarrow \text{CH}_3\text{OH}(l) + \tfrac32\,\text{O}_2(g) \quad \Delta H = +726$

   5. Add them up — CO₂ and H₂O cancel, and O₂ cancels as well (check: left side O₂ from C combustion + from H₂ combustion = 1 + 1 = 2 O₂; right side has 1.5 O₂; net left has 0.5 O₂, which matches the formation reaction). Sum of enthalpies:

   $\Delta H_f^\circ = (-393) + (-572) + (+726) = -239\ \text{kJ/mol}$

   Tip

   A quick way: formation enthalpy = sum of combustion enthalpies of elements (each multiplied by their coefficient) minus combustion enthalpy of the compound. Here:

   $\Delta H_f^\circ(\text{CH}_3\text{OH}) = [1 \times (-393) + 2 \times (-286)] - (-726) = -239$

   Watch out

   A common mistake is forgetting to multiply the H₂ combustion by 2, or adding instead of subtracting the methanol combustion. Always check that the stoichiometry of O₂ balances.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

10. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.Enthalpy of formation of CO(g), CO$_2$(g) are $-110$, $-393$ kJmol$^{-1}$ respectively. The enthalpy of combustion of CO (in kJmol$^{-1}$) is (A) $-283.0$ (B) $-110.5$ (C) $504$ (D) $-221.2$

    ›Reveal solutionSolution

    The enthalpy of combustion of CO is found by applying Hess’s law: the combustion reaction is CO + ½O₂ → CO₂, and its enthalpy equals ΔH_f(CO₂) − ΔH_f(CO) = −393 − (−110) = −283 kJ mol⁻¹, so the correct option is (A).

    The key idea is that enthalpy of combustion is just the enthalpy change when one mole of a substance burns completely in oxygen. For CO, the combustion reaction is:

    $\text{CO(g)} + \tfrac12 \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$

    We are given the standard enthalpies of formation:

    • ΔH_f°(CO) = −110 kJ mol⁻¹
    • ΔH_f°(CO₂) = −393 kJ mol⁻¹

    Since enthalpy is a state function, we can use Hess’s law: the enthalpy change for a reaction equals the sum of the enthalpies of formation of products minus the sum for reactants.

    1. Write the target reaction

    $\text{CO(g)} + \tfrac12 \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$

    2. Apply the formation enthalpy formula For any reaction:

    $\Delta H_{\text{rxn}} = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$

    Here, the only product is CO₂, and the only reactant with a non-zero formation enthalpy is CO (O₂ is an element in its standard state, so its ΔH_f° = 0).

    3. Plug in the numbers

    $\Delta H_{\text{comb}} = \Delta H_f^\circ(\text{CO}_2) - \Delta H_f^\circ(\text{CO})$

    $\Delta H_{\text{comb}} = (-393) - (-110) = -393 + 110 = -283 \text{ kJ mol}^{-1}$

    4. Interpret the sign The negative value confirms that combustion of CO is exothermic, as expected.

    Watch out

    A common mistake is to forget that O₂ has ΔH_f° = 0, or to accidentally subtract in the wrong order. Always remember: products minus reactants.

    Tip

    You can also think of this as a “thermochemical cycle”: forming CO₂ directly from elements gives −393 kJ, but forming CO from elements first costs −110 kJ (releases less heat), so burning CO to CO₂ must release the difference: 393 − 110 = 283 kJ.

    ✓Final answer

    The correct option is (A).

    ANSWER: A

11. TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
    Q.At $27^\circ \mathrm{C}$, the osmotic pressure of $0.5\,\mathrm{M}$ solution of sucrose (in bar) is $(R = 0.083\,\mathrm{L\,bar\,mol^{-1}\,K^{-1}})$ (A) $12.45$ (B) $24.90$ (C) $84.63$ (D) $7.61$

    ›Reveal solutionSolution

    Osmotic pressure is a colligative property given by $\pi = iCRT$. For a non-electrolyte like sucrose, $i = 1$, so $\pi = CRT$. At $27^\circ \mathrm{C}$ ($300\,\mathrm{K}$) with $C = 0.5\,\mathrm{M}$ and $R = 0.083\,\mathrm{L\,bar\,mol^{-1}\,K^{-1}}$, the osmotic pressure is $12.45\,\mathrm{bar}$, which corresponds to option (A).

    Osmotic pressure is one of the four colligative properties — properties that depend only on the number of solute particles, not on their identity. The key formula is $\pi = iCRT$, where $i$ is the van't Hoff factor (number of particles per formula unit in solution). Sucrose is a non-electrolyte; it dissolves as intact molecules, so $i = 1$. That means the formula simplifies directly to $\pi = CRT$.

    The temperature must be in Kelvin. $27^\circ \mathrm{C}$ converts to $T = 27 + 273 = 300\,\mathrm{K}$. The concentration is given as $0.5\,\mathrm{M}$, which means $0.5$ moles per litre. The gas constant $R$ is provided in units of $\mathrm{L\,bar\,mol^{-1}\,K^{-1}}$, so the answer will come out directly in bar — no unit conversion needed.

    1. Write the formula: $\pi = CRT$.
    2. Substitute the values: $C = 0.5$, $R = 0.083$, $T = 300$.
    3. Multiply: $\pi = 0.5 \times 0.083 \times 300$.
    4. First, $0.5 \times 300 = 150$.
    5. Then, $150 \times 0.083 = 12.45$.

    The calculation is straightforward: $150 \times 0.083 = 150 \times \frac{83}{1000} = \frac{12450}{1000} = 12.45$.

    Watch out

    A common mistake is to forget to convert Celsius to Kelvin. Using $27^\circ \mathrm{C}$ directly gives $0.5 \times 0.083 \times 27 = 1.1205$, which is not among the options — but it's dangerously close to a wrong choice if the numbers were different. Always add 273.

    Tip

    Notice that $0.083 \times 300 = 24.9$, and half of that is $12.45$. So you can think: $\pi = \frac{1}{2} \times (0.083 \times 300) = \frac{24.9}{2} = 12.45$. This mental shortcut saves time in exams.

    ✓Final answer

    The osmotic pressure is $\boxed{12.45\,\mathrm{bar}}$, which corresponds to option (A).

12. TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQ
    Q.The equilibrium constant $K_c$ of the following reaction is 9 at a temperature T. A(g) + B(g) $\rightleftharpoons$ C(g) + D(g) Assuming that the initial concentration of all the gases is 1 mole, what will be the concentration of A and C, respectively, at equilibrium? (A) 0.5 ; 1.5 (B) $\left(1 - \dfrac{3}{\sqrt{10}}\right) ; \left(1 + \dfrac{3}{\sqrt{10}}\right)$ (C) 0.5 ; 0.5 (D) 1.5 ; 1.5

    ›Reveal solutionSolution

    The symmetric equilibrium ($K_c = 9$, all start at 1) gives $x = 0.5$, so $[A] = 0.5$ and $[C] = 1.5$ — option (A).

    Idea. Equal moles on both sides mean the total moles are unchanged, so we work directly with concentrations. Let $x$ react.

    ICE table. Start: $[A]_0=[B]_0=[C]_0=[D]_0 = 1$ M. Let $x$ of A and B react:

    $[A] = 1-x,\quad [B] = 1-x,\quad [C] = 1+x,\quad [D] = 1+x.$

    Equilibrium expression.

    $K_c = \frac{[C][D]}{[A][B]} = \frac{(1+x)^2}{(1-x)^2} = 9.$

    Solve. Take the positive square root (all concentrations positive for $0<x<1$):

    $\frac{1+x}{1-x} = 3 \;\Rightarrow\; 1+x = 3(1-x) \;\Rightarrow\; 1+x = 3-3x \;\Rightarrow\; 4x = 2 \;\Rightarrow\; x = 0.5.$

    Concentrations.

    $[A] = 1 - 0.5 = 0.5\ \text{M}, \qquad [C] = 1 + 0.5 = 1.5\ \text{M}.$

    Check. $K_c = \dfrac{(1.5)^2}{(0.5)^2} = \dfrac{2.25}{0.25} = 9$. Consistent.

    Watch out

    Because the reactant and product ICE terms are perfect squares, the equilibrium expression reduces to $\dfrac{1+x}{1-x}=\sqrt{K_c}$ — a linear equation, so no messy quadratic is needed.

    ✓Final answer

    $[A] = 0.5$ M and $[C] = 1.5$ M. The correct option is (A).

    ANSWER: A