NCERT Solutions · Class 11 Chemistry Thermodynamics

An adiabatic process is defined by zero heat exchange with the surroundings — not by zero temperature change, zero pressure change, or zero work.Start from the first law of thermodynamics, \[\Delta U = q + w \] where \(\displaystyle \Delta U\) is the change in internal energy of the system, \(\displaystyle q\) is the heat absorbed by the system from its surroundings, and \(\displaystyle w\) is the work done on the system.The word "adiabatic" refers specifically to the heat term: an adiabatic process is one carried out with the system thermally insulated from its surroundings, so no heat can enter or leave it. That condition is written as \[q = 0 \]Now test the other three options against this definition, one at a time. The step students most often get wrong: "adiabatic" is a statement about heat (\(\displaystyle q\)), while "no work done" is a separate constraint tied to volume, and the two must not be swapped. A process can be adiabatic and still involve a large amount of work — that is exactly how an adiabatic expansion cools a gas.Only option (iii) matches the definition of an adiabatic process.Answer: (iii) q = $\displaystyle 0$ — an adiabatic process is one in which no heat is exchanged between the system and its surroundings.

Standard enthalpy of formation of an element in its own standard state is fixed at zero — it is a reference point, not something measured or calculated for each element.Enthalpy \(\displaystyle H \) itself can never be measured in an absolute sense — only changes in enthalpy, \(\displaystyle \Delta H \), between two states can be measured. So to build a usable scale of enthalpies for compounds, chemists need a common reference (a "sea level") from which every other enthalpy is measured.That reference is the standard enthalpy of formation, \(\displaystyle \Delta_f H^\ominus \): the enthalpy change when $\displaystyle 1$ mole of a compound is formed from its elements, with every element in its most stable form at the standard state ($\displaystyle 298$ K, $\displaystyle 1$ bar).By this very definition, forming an element from itself involves no chemical change at all — nothing is transformed, nothing is broken or made — so there is no enthalpy change to record: \[\Delta_f H^\ominus(\text{element, standard state}) = 0 \]This is a convention fixed by IUPAC, the same way sea level is arbitrarily set to $\displaystyle 0$ m in geography even though the ocean has a real depth beneath it. It does not mean elements contain "no energy" — it only means their enthalpy is the zero mark against which every compound's enthalpy of formation is measured.A common slip is to think this value should depend on the element (heavier atoms, more electrons, more energy stored), and pick option (iv). But the zero is not a statement about the physical energy content of the element — it is a bookkeeping choice, identical for every element, made purely so that \(\displaystyle \Delta_f H^\ominus \) of compounds can be compared on one common scale.So of the four choices, "unity" (i) and "less than zero" (iii) are not the convention at all, and "different for each element" (iv) misunderstands why the reference is chosen — it must be the same value (zero) for every element to work as a common baseline.Answer: (ii) zero — the standard enthalpy of every element in its standard state is taken as zero by convention, since it forms the reference point from which all other standard enthalpies (of formation) are measured.

At constant pressure, \(\displaystyle \Delta H \) and \(\displaystyle \Delta U \) differ by the work done by gas moles that appear or disappear — not by chance, but by a fixed formula.For a reaction run at constant temperature, the relation between enthalpy change and internal energy change is \[\Delta H = \Delta U + \Delta n_g RT \] where \(\displaystyle \Delta n_g \) is the change in the number of moles of gaseous species (moles of gaseous products − moles of gaseous reactants), \(\displaystyle R \) is the gas constant, and \(\displaystyle T \) is the absolute temperature. This formula comes from \(\displaystyle H = U + pV \) and treating the gases as ideal, so \(\displaystyle pV = n_g RT \).Write the balanced combustion equation and note the physical states — this is the step people skip. \[\mathrm{CH_4(g) + 2O_2(g) \longrightarrow CO_2(g) + 2H_2O(l)} \] Water is produced as a liquid at standard conditions, not a gas — only \(\displaystyle \mathrm{CH_4}, \mathrm{O_2} \), and \(\displaystyle \mathrm{CO_2} \) count toward \(\displaystyle \Delta n_g \).Gaseous moles on the product side: \(\displaystyle 1 \) (from \(\displaystyle \mathrm{CO_2} \)) Gaseous moles on the reactant side: \(\displaystyle 1 + 2 = 3 \) (from \(\displaystyle \mathrm{CH_4} \) and \(\displaystyle \mathrm{O_2} \))\[\Delta n_g = 1 - 3 = -2 \]Substitute \(\displaystyle \Delta n_g \) into the formula. \[\Delta H^0 = \Delta U^0 + (-2)RT = \Delta U^0 - 2RT \]Since \(\displaystyle R \) and \(\displaystyle T \) are both positive, \(\displaystyle 2RT \) is a positive quantity being subtracted. That means \[\Delta H^0 < \Delta U^0 \] regardless of the numerical value of \(\displaystyle X \). (To see the size of the correction: at \(\displaystyle T = 298\ \mathrm{K} \), \(\displaystyle 2RT = 2 \times 8.314 \times 10^{-3}\ \mathrm{kJ\ mol^{-1}K^{-1}} \times 298\ \mathrm{K} \approx 4.96\ \mathrm{kJ\ mol^{-1}} \) — a small but definite shift, always in the direction that makes \(\displaystyle \Delta H^0 \) more negative than \(\displaystyle \Delta U^0 \) here.)The combustion is exothermic, so \(\displaystyle \Delta U^0 = -X\ \mathrm{kJ\ mol^{-1}} \) is already negative; subtracting a further positive number (\(\displaystyle 2RT \)) makes \(\displaystyle \Delta H^0 \) even more negative, confirming \(\displaystyle \Delta H^0 < \Delta U^0 \).Answer: (iii) \(\displaystyle \Delta H^0 < \Delta U^0 \), because \(\displaystyle \Delta n_g = -2 \) for this combustion ($\displaystyle 3$ mol gas → $\displaystyle 1$ mol gas, since water forms as a liquid), giving \(\displaystyle \Delta H^0 = \Delta U^0 - 2RT \).

Hess's Law says enthalpy is a state function — so instead of measuring the formation reaction directly, you can build it by combining reactions you already have data for, as long as the reactants and products add up correctly.The target reaction is the formation of methane from its elements in their standard states:\[\text{C(graphite)} + 2\text{H}_2(g) \rightarrow \text{CH}_4(g), \qquad \Delta_fH^\circ = \, ? \]You are given three combustion reactions instead:\[\text{(1) } \text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l), \qquad \Delta_cH_1 = -890.3\ \text{kJ mol}^{-1} \] \[\text{(2) } \text{C(graphite)} + \text{O}_2(g) \rightarrow \text{CO}_2(g), \qquad \Delta_cH_2 = -393.5\ \text{kJ mol}^{-1} \] \[\text{(3) } \text{H}_2(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l), \qquad \Delta_cH_3 = -285.8\ \text{kJ mol}^{-1} \]The trick people miss: CH₄ is a product of the reaction you want, but a reactant in equation ($\displaystyle 1$) — so equation ($\displaystyle 1$) must be reversed (its sign flipped) when you add it in.Build the target by combining: reaction ($\displaystyle 2$), plus twice reaction ($\displaystyle 3$) (to get \(\displaystyle 2\text{H}_2\) instead of one), minus reaction ($\displaystyle 1$) (to flip CH₄ from reactant to product):\[\underbrace{\text{C} + \text{O}_2 \rightarrow \text{CO}_2}_{(2)} \;+\; 2\underbrace{\left(\text{H}_2 + \tfrac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O}\right)}_{(3)} \;-\; \underbrace{\left(\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}\right)}_{(1)} \]Add up the species on each side: \(\displaystyle \text{CO}_2\) appears on both sides and cancels, \(\displaystyle 2\text{H}_2\text{O}\) appears on both sides and cancels, and the \(\displaystyle \text{O}_2\) totals (\(\displaystyle 1 + 2\times\tfrac{1}{2} = 2\) on the left of the sum, \(\displaystyle 2\) from reaction ($\displaystyle 1$) moved to the left) cancel too. What survives is exactly\[\text{C(graphite)} + 2\text{H}_2(g) \rightarrow \text{CH}_4(g) \]which is the formation reaction you want. Since enthalpy changes add the same way the equations do, this fixes the formula:\[\Delta_fH^\circ(\text{CH}_4) = \Delta_cH_2 + 2\Delta_cH_3 - \Delta_cH_1 \]Substituting the values (all in kJ mol⁻¹, so the units carry straight through without conversion):\[\Delta_fH^\circ(\text{CH}_4) = (-393.5) + 2(-285.8) - (-890.3) \]\[\Delta_fH^\circ(\text{CH}_4) = -393.5 - 571.6 + 890.3 \]\[\Delta_fH^\circ(\text{CH}_4) = -74.8\ \text{kJ mol}^{-1} \]The result is negative, which makes physical sense: forming a stable molecule like methane from its elements releases energy, exactly as combustion (also exothermic) would suggest.Answer: \(\displaystyle \Delta_fH^\circ(\text{CH}_4) = -74.8\ \text{kJ mol}^{-1}\), option (i).

Spontaneity is decided by \(\displaystyle \Delta G \), not by \(\displaystyle \Delta S \) alone.The condition for a reaction to be spontaneous ("possible") at a given temperature is the Gibbs free energy equation\[\Delta G = \Delta H - T\Delta S \]where \(\displaystyle \Delta G \) is the free energy change, \(\displaystyle \Delta H \) is the enthalpy change, \(\displaystyle T \) is the absolute temperature (in kelvin), and \(\displaystyle \Delta S \) is the entropy change. A reaction runs on its own only when \(\displaystyle \Delta G < 0 \).Read the equation for what \(\displaystyle q \) means. The reaction is written as\[A + B \rightarrow C + D + q \]Heat \(\displaystyle q \) appears on the product side, which is the standard way of saying heat is released by the reaction — this is an exothermic reaction, so\[\Delta H < 0 \]The step people skip is exactly this: a "+q" tacked onto the products is a statement about \(\displaystyle \Delta H \), and mistaking it for something about \(\displaystyle \Delta S \) is what leads to the wrong option here.Now put both signs into the Gibbs equation. We are told separately that the entropy change is positive:\[\Delta S > 0 \]So \(\displaystyle -T\Delta S \) is a negative quantity at every temperature \(\displaystyle T > 0 \) (since \(\displaystyle T \) is always positive on the Kelvin scale). Substituting the two signs:\[\Delta G = \underbrace{\Delta H}_{\text{negative}} - \underbrace{T\Delta S}_{\text{positive}} = (\text{negative}) + (\text{negative}) \]Both terms on the right push \(\displaystyle \Delta G \) in the same direction — there is no temperature at which they can cancel to make \(\displaystyle \Delta G \) positive. Hence\[\Delta G < 0 \quad \text{for every value of } T \]Contrast with the cases that do depend on temperature. A reaction is only temperature-sensitive (spontaneous at high \(\displaystyle T \) but not low \(\displaystyle T \), or the reverse) when \(\displaystyle \Delta H \) and \(\displaystyle \Delta S \) have opposite signs, because then \(\displaystyle \Delta H \) and \(\displaystyle -T\Delta S \) compete. Here they don't compete — exothermic (\(\displaystyle \Delta H < 0 \)) combined with increasing disorder (\(\displaystyle \Delta S > 0 \)) makes \(\displaystyle \Delta G \) negative unconditionally, so raising or lowering \(\displaystyle T \) can never flip the sign.Answer: the reaction is possible at any temperature (exothermic with \(\displaystyle \Delta S > 0 \) gives \(\displaystyle \Delta G = \Delta H - T\Delta S < 0 \) for all \(\displaystyle T \), since both terms are negative).

Change in internal energy comes from the First Law: heat absorbed adds to it, work done BY the system subtracts from it.The First Law of Thermodynamics states \[\Delta U = q + w \] where \(\displaystyle \Delta U \) is the change in internal energy of the system, \(\displaystyle q \) is the heat absorbed by the system, and \(\displaystyle w \) is the work done on the system.The sign convention is the step people trip on: heat absorbed by the system is positive, and work done by the system (i.e., the system pushing on its surroundings) is negative, because that energy leaves the system.Here the system absorbs heat, so \[q = +701\ \text{J} \]The system does work on the surroundings, so that energy is lost from the system: \[w = -394\ \text{J} \]Substituting into the First Law: \[\Delta U = q + w = (+701\ \text{J}) + (-394\ \text{J}) \]\[\Delta U = 701\ \text{J} - 394\ \text{J} = 307\ \text{J} \]Both given quantities ($\displaystyle 701$ J, $\displaystyle 394$ J) are whole numbers of joules, so the result is reported to the same precision, with no decimal places.Answer: \(\displaystyle \Delta U = 307\ \text{J} \) (the internal energy of the system increases by $\displaystyle 307$ J).

\(\displaystyle \Delta H = \Delta U + \Delta n_g RT\), and this whole question is the sign of \(\displaystyle \Delta n_g\) — which counts GASES only, so a solid reactant and a liquid product contribute nothing.The bomb calorimeter measures at constant volume, so what it gives you is \(\displaystyle \Delta U\), not \(\displaystyle \Delta H\). Converting between them needs the change in the number of moles of GAS:\[\Delta H = \Delta U + (\Delta n_g) RT \]where \(\displaystyle \Delta n_g\) = (moles of gaseous products) − (moles of gaseous reactants), \(\displaystyle R = 8.314\ \mathrm{J\,K^{-1}mol^{-1}}\) and \(\displaystyle T = 298\ \mathrm{K}\), so\[RT = \frac{8.314 \times 298}{1000} = 2.478\ \mathrm{kJ\,mol^{-1}} \]Counting the gases. On the product side: \(\displaystyle \mathrm{N_2(g)}\) and \(\displaystyle \mathrm{CO_2(g)}\) give $\displaystyle 2$ moles of gas. \(\displaystyle \mathrm{H_2O(l)}\) is a LIQUID and contributes nothing — this is the step people lose the mark on. On the reactant side, \(\displaystyle \tfrac{3}{2}\,\mathrm{O_2(g)}\) gives $\displaystyle 1.5$ moles of gas, and cyanamide contributes $\displaystyle 1$ mole if it is a gas and nothing if it is a solid.This exercise labels the cyanamide two different ways. Its prose calls it \(\displaystyle \mathrm{NH_2CN\,(s)}\); the equation printed underneath writes \(\displaystyle \mathrm{NH_2CN(g)}\). The two readings give opposite signs for \(\displaystyle \Delta n_g\), so work both and know which you are being asked for.Reading it as the solid (which is what cyanamide is at $\displaystyle 298$ K — it melts at about $\displaystyle 44$ °C, and a bomb calorimeter burns a weighed solid pellet):\[\Delta n_g = 2 - 1.5 = +0.5 \] \[\Delta H = -742.7 + (0.5)(2.478) = -742.7 + 1.239 = -741.5\ \mathrm{kJ\,mol^{-1}} \]Reading it as the gas, as the printed equation writes it:\[\Delta n_g = 2 - (1 + 1.5) = -0.5 \] \[\Delta H = -742.7 - 1.239 = -743.9\ \mathrm{kJ\,mol^{-1}} \]NCERT's answer key prints \(\displaystyle -743.939\ \mathrm{kJ}\), so the key follows the equation as printed rather than the prose. Nothing else in the calculation moves: only the \(\displaystyle 1.239\ \mathrm{kJ}\) term changes sign.In an exam, answer to the state symbols on the paper in front of you, and show the \(\displaystyle \Delta n_g\) line either way — that line is where the marks are, not the third decimal place.Answer: \(\displaystyle \Delta H = -741.5\ \mathrm{kJ\,mol^{-1}}\), taking \(\displaystyle \mathrm{NH_2CN}\) as the solid it is at $\displaystyle 298$ K. NCERT's key prints \(\displaystyle -743.939\ \mathrm{kJ}\), which is the same calculation with the cyanamide read as a gas, as the exercise's own printed equation writes it.

Heat absorbed at constant pressure is \(\displaystyle q = n \, C_m \, \Delta T \), where \(\displaystyle n\) is the number of moles, \(\displaystyle C_m\) is the molar heat capacity, and \(\displaystyle \Delta T\) is the temperature change. You need moles of aluminium, not grams, because \(\displaystyle C_m\) is defined per mole.Step $\displaystyle 1$: Convert mass to moles.The molar mass of aluminium is \(\displaystyle M = 27.0\ \text{g mol}^{-1}\).\[n = \frac{\text{mass}}{M} = \frac{60.0\ \text{g}}{27.0\ \text{g mol}^{-1}} = 2.222\ \text{mol} \]Step $\displaystyle 2$: Find the temperature change.\[\Delta T = 55\,^{\circ}\text{C} - 35\,^{\circ}\text{C} = 20\,^{\circ}\text{C} = 20\ \text{K} \]A change of \(\displaystyle 20\,^{\circ}\text{C}\) equals a change of \(\displaystyle 20\ \text{K}\) — the two scales differ only by the fixed $\displaystyle 273.15$ offset, which cancels out in a difference. This is different from converting a single temperature reading, where you must add 273.Step $\displaystyle 3$: Apply \(\displaystyle q = n \, C_m \, \Delta T \).Here \(\displaystyle C_m = 24\ \text{J mol}^{-1}\ \text{K}^{-1}\) is the molar heat capacity of Al (heat needed to raise $\displaystyle 1$ mole by $\displaystyle 1$ K).\[q = (2.222\ \text{mol}) \times (24\ \text{J mol}^{-1}\text{K}^{-1}) \times (20\ \text{K}) \]\[q = 2.222 \times 24 \times 20\ \text{J} = 1066.7\ \text{J} \]Step $\displaystyle 4$: Convert to kJ.\[q = 1066.7\ \text{J} \times \frac{1\ \text{kJ}}{1000\ \text{J}} = 1.0667\ \text{kJ} \]The given data ($\displaystyle 60.0$ g, $\displaystyle 35$ °C, $\displaystyle 55$ °C, $\displaystyle 24$ J mol⁻¹ K⁻¹) each carry two to three significant figures, so the result is rounded once, at the end, to three significant figures.Answer: \(\displaystyle q \approx 1.07\ \text{kJ}\)

Enthalpy is a state function, so you can break an odd path (liquid at +$\displaystyle 10$ °C → solid at –$\displaystyle 10$ °C) into a chain of simple steps — cooling, then freezing, then cooling again — and just add up the \(\displaystyle \Delta H\) of each step, even though water never actually stops at $\displaystyle 0$ °C on its way down.The direct process — water at $\displaystyle 10.0$ °C freezing to ice at –$\displaystyle 10.0$ °C — is not something you have data for. But Hess's law says the total enthalpy change depends only on the start and end states, not the route. So build a three-step route between the same two states, using only the numbers you're given:Step $\displaystyle 1$: cool liquid water from $\displaystyle 10.0$ °C to $\displaystyle 0.0$ °C (no phase change) Step $\displaystyle 2$: freeze the liquid to ice, at $\displaystyle 0.0$ °C (the only temperature you have \(\displaystyle \Delta_{fus}H\) for) Step $\displaystyle 3$: cool the ice from $\displaystyle 0.0$ °C to –$\displaystyle 10.0$ °C (no phase change)Step $\displaystyle 1$ — cooling the liquidFor a temperature change with no phase change, use \(\displaystyle q_p = n C_p \Delta T\), where \(\displaystyle n\) is the number of moles, \(\displaystyle C_p\) is the molar heat capacity at constant pressure, and \(\displaystyle \Delta T = T_{final} - T_{initial}\).\[\Delta H_1 = n\,C_p[\text{H}_2\text{O}(l)]\,\Delta T = (1.0\ \text{mol})(75.3\ \text{J mol}^{-1}\text{K}^{-1})(0.0^\circ\text{C} - 10.0^\circ\text{C}) \]Since \(\displaystyle \Delta T\) is a difference of temperatures, it's the same whether you read it in °C or in K — the $\displaystyle 273.15$ cancels out. So:\[\Delta H_1 = (1.0)(75.3)(-10.0)\ \text{J} = -753\ \text{J} = -0.753\ \text{kJ} \]Step $\displaystyle 2$ — freezing at $\displaystyle 0$ °CYou're given \(\displaystyle \Delta_{fus}H = 6.03\ \text{kJ mol}^{-1}\) — that's for melting (ice → liquid). Freezing is the reverse process, so its enthalpy change is the same size with the opposite sign. This is the step people skip: forgetting to flip the sign for the reverse of fusion gives an answer with the wrong direction of heat flow.\[\Delta H_2 = -\Delta_{fus}H = -6.03\ \text{kJ mol}^{-1} \times 1.0\ \text{mol} = -6.03\ \text{kJ} \]Step $\displaystyle 3$ — cooling the iceSame formula as Step $\displaystyle 1$, now with the solid's heat capacity:\[\Delta H_3 = n\,C_p[\text{H}_2\text{O}(s)]\,\Delta T = (1.0\ \text{mol})(36.8\ \text{J mol}^{-1}\text{K}^{-1})(-10.0^\circ\text{C} - 0.0^\circ\text{C}) \]\[\Delta H_3 = (1.0)(36.8)(-10.0)\ \text{J} = -368\ \text{J} = -0.368\ \text{kJ} \]Adding the three stepsBecause enthalpy is a state function, the sum of these three step-changes equals the enthalpy change for the actual, direct process:\[\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = (-0.753\ \text{kJ}) + (-6.03\ \text{kJ}) + (-0.368\ \text{kJ}) \]\[\Delta H = -7.151\ \text{kJ} \]Every piece of data here carries three significant figures ($\displaystyle 6.03$, $\displaystyle 75.3$, $\displaystyle 36.8$, $\displaystyle 10.0$), so the final value is rounded to three significant figures too:\[\Delta H \approx -7.15\ \text{kJ mol}^{-1} \]The negative sign makes physical sense: freezing releases heat (it's exothermic), and cooling the water down before and after also releases heat, so all three steps push in the same direction.Answer: \(\displaystyle \Delta H = -7.15\ \text{kJ mol}^{-1}\)