NCERT Solutions · Class 11 Chemistry Thermodynamics

Enthalpy of combustion tells you the heat released per mole of substance formed — not per gram. To use it for $\displaystyle 35.2$ g of \(\displaystyle \text{CO}_2\), first convert that mass into moles.The reaction is \[\text{C(s)} + \text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}), \qquad \Delta_cH = -393.5\ \text{kJ mol}^{-1} \]The negative sign means the reaction is exothermic — $\displaystyle 393.5$ kJ of heat is released for every $\displaystyle 1$ mole of \(\displaystyle \text{CO}_2\) formed.Step $\displaystyle 1$: Convert mass of \(\displaystyle \text{CO}_2\) to moles.Number of moles, \(\displaystyle n = \dfrac{\text{given mass}}{\text{molar mass}}\), where the molar mass of \(\displaystyle \text{CO}_2\) is \[M(\text{CO}_2) = 12 + 2(16) = 44\ \text{g mol}^{-1} \]This is the step people rush past: you cannot plug $\displaystyle 35.2$ g directly into an enthalpy given "per mole" — the unit mismatch has to be fixed first.\[n(\text{CO}_2) = \frac{35.2\ \text{g}}{44\ \text{g mol}^{-1}} = 0.8\ \text{mol} \]Step $\displaystyle 2$: Scale the enthalpy of combustion to this many moles.Heat released, \(\displaystyle q = n \times |\Delta_cH|\), where \(\displaystyle n\) is the moles of \(\displaystyle \text{CO}_2\) formed and \(\displaystyle \Delta_cH\) is the enthalpy of combustion per mole.\[q = 0.8\ \text{mol} \times 393.5\ \text{kJ mol}^{-1} \]\[q = 314.8\ \text{kJ} \]The data ($\displaystyle 393.5$, four significant figures; $\displaystyle 35.2$ g, three significant figures) justifies keeping three significant figures in the final answer, so \(\displaystyle q = 315\ \text{kJ}\).Since combustion is exothermic, this heat is released (given out), consistent with the negative sign on \(\displaystyle \Delta_cH\).Answer: $\displaystyle 315$ kJ ($\displaystyle 314.8$ kJ) of heat is released.

Use Hess's law: \(\displaystyle \Delta_r H = \Sigma \Delta_f H(\text{products}) - \Sigma \Delta_f H(\text{reactants}) \), each term weighted by its coefficient in the balanced equation.Here \(\displaystyle \Delta_f H \) is the standard enthalpy of formation of each substance — the enthalpy change when $\displaystyle 1$ mole of that compound forms from its elements in their standard states. This law works because enthalpy is a state function: it doesn't matter whether the reactants go straight to products or are imagined to pass through their elements first, so building products from elements and tearing reactants down to elements (in the reverse direction) must sum to the same \(\displaystyle \Delta_r H \) as the direct reaction.The reaction is\[\text{N}_2\text{O}_4(g) + 3\text{CO}(g) \rightarrow \text{N}_2\text{O}(g) + 3\text{CO}_2(g) \]The given formation enthalpies are\[\Delta_f H[\text{CO}(g)] = -110 \text{ kJ mol}^{-1}, \quad \Delta_f H[\text{CO}_2(g)] = -393 \text{ kJ mol}^{-1} \] \[\Delta_f H[\text{N}_2\text{O}(g)] = 81 \text{ kJ mol}^{-1}, \quad \Delta_f H[\text{N}_2\text{O}_4(g)] = 9.7 \text{ kJ mol}^{-1} \]The step people get wrong here: every formation enthalpy must be multiplied by its own stoichiometric coefficient from the balanced equation — CO and CO₂ each carry a factor of $\displaystyle 3$ — before you subtract. Skipping this scaling is the most common way to get this type of question wrong.Applying Hess's law:\[\Delta_r H = \Big[1 \times \Delta_f H(\text{N}_2\text{O}) + 3 \times \Delta_f H(\text{CO}_2)\Big] - \Big[1 \times \Delta_f H(\text{N}_2\text{O}_4) + 3 \times \Delta_f H(\text{CO})\Big] \]Substitute the values, keeping units of kJ mol⁻¹ attached throughout:\[\Delta_r H = \Big[(81 \text{ kJ mol}^{-1}) + 3 \times (-393 \text{ kJ mol}^{-1})\Big] - \Big[(9.7 \text{ kJ mol}^{-1}) + 3 \times (-110 \text{ kJ mol}^{-1})\Big] \]Work out each bracket separately.Products bracket:\[81 \text{ kJ mol}^{-1} + (-1179 \text{ kJ mol}^{-1}) = -1098 \text{ kJ mol}^{-1} \]Reactants bracket:\[9.7 \text{ kJ mol}^{-1} + (-330 \text{ kJ mol}^{-1}) = -320.3 \text{ kJ mol}^{-1} \]Now subtract reactants from products:\[\Delta_r H = -1098 \text{ kJ mol}^{-1} - (-320.3 \text{ kJ mol}^{-1}) = -1098 \text{ kJ mol}^{-1} + 320.3 \text{ kJ mol}^{-1} \]\[\Delta_r H = -777.7 \text{ kJ mol}^{-1} \]The data are given to one decimal place at best ($\displaystyle 9.7$ kJ mol⁻¹), so the result is reported to one decimal place, matching the input precision — no further rounding is needed since the arithmetic above is exact.The negative sign shows the reaction releases heat: N₂O₄ and CO together are higher in enthalpy than the N₂O and CO₂ they form, so energy comes out as the bonds rearrange.Answer: \(\displaystyle \Delta_r H = -777.7 \text{ kJ mol}^{-1} \)

Standard enthalpy of formation is defined per ONE mole of the compound formed from its elements in their standard states — but the equation given here makes $\displaystyle 2$ mol of \(\displaystyle \text{NH}_3\), so you must divide before you can call the number a formation enthalpy.The given thermochemical equation is\[\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g), \qquad \Delta_r H^0 = -92.4\ \text{kJ mol}^{-1} \]Here \(\displaystyle \Delta_r H^0\) is the enthalpy change for the reaction exactly as written — that is, for the formation of $\displaystyle 2$ mol of \(\displaystyle \text{NH}_3(g)\) from $\displaystyle 1$ mol \(\displaystyle \text{N}_2(g)\) and $\displaystyle 3$ mol \(\displaystyle \text{H}_2(g)\), both elements already in their standard states.The standard enthalpy of formation, \(\displaystyle \Delta_f H^0\), of a substance is the enthalpy change when $\displaystyle 1$ mol of that substance is formed from its elements in their standard states. So:\[\Delta_f H^0(\text{NH}_3) = \frac{\Delta_r H^0}{\text{moles of NH}_3 \text{ formed}} \]This is the step people skip: they copy \(\displaystyle -92.4\ \text{kJ mol}^{-1}\) straight in as the formation enthalpy, but that number belongs to the reaction producing $\displaystyle 2$ mol of product, not to $\displaystyle 1$ mol of \(\displaystyle \text{NH}_3\).Substituting the values:\[\Delta_f H^0(\text{NH}_3) = \frac{-92.4\ \text{kJ}}{2\ \text{mol}} \]\[\Delta_f H^0(\text{NH}_3) = -46.2\ \text{kJ mol}^{-1} \]The data (\(\displaystyle -92.4\)) is given to $\displaystyle 3$ significant figures, so the halved result is kept to the same precision: \(\displaystyle -46.2\ \text{kJ mol}^{-1}\), with no further rounding needed since the division is exact.Answer: \(\displaystyle \Delta_f H^0(\text{NH}_3) = -46.2\ \text{kJ mol}^{-1}\)

Hess's Law: enthalpy change depends only on the initial and final states, not on the path. So the enthalpy of formation of \(\displaystyle \mathrm{CH_3OH(l)} \) can be built by combining the three given equations, even though none of them is that formation reaction by itself.The target reaction — the standard enthalpy of formation of methanol — is the one where $\displaystyle 1$ mole of \(\displaystyle \mathrm{CH_3OH(l)} \) is formed from its elements in their standard states: \[\mathrm{C(graphite) + 2H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow CH_3OH(l)}, \qquad \Delta_f H^0 = \, ? \]The three data equations are: \[\text{(i)} \quad \mathrm{CH_3OH(l) + \tfrac{3}{2}O_2(g) \rightarrow CO_2(g) + 2H_2O(l)}, \quad \Delta_r H^0_{(i)} = -726 \text{ kJ mol}^{-1} \] \[\text{(ii)} \quad \mathrm{C(graphite) + O_2(g) \rightarrow CO_2(g)}, \quad \Delta_c H^0_{(ii)} = -393 \text{ kJ mol}^{-1} \] \[\text{(iii)} \quad \mathrm{H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow H_2O(l)}, \quad \Delta_f H^0_{(iii)} = -286 \text{ kJ mol}^{-1} \]Finding the combination is bookkeeping on atoms, done before touching a single number. The target has \(\displaystyle \mathrm{CH_3OH(l)} \) as a product, but in (i) it is a reactant — so (i) must be reversed (flip the sign of its \(\displaystyle \Delta H \)). The target needs $\displaystyle 2$ mol of \(\displaystyle \mathrm{H_2}\), and (iii) supplies $\displaystyle 1$ mol of \(\displaystyle \mathrm{H_2}\) — so (iii) is used twice. Equation (ii) already matches the carbon term as written, used once.Check this combination reproduces the target: reverse (i), add (ii), add \(\displaystyle 2\times \)(iii):\[\mathrm{CO_2(g) + 2H_2O(l) \rightarrow CH_3OH(l) + \tfrac{3}{2}O_2(g)} \] \[+\ \mathrm{C(graphite) + O_2(g) \rightarrow CO_2(g)} \] \[+\ \mathrm{2H_2(g) + O_2(g) \rightarrow 2H_2O(l)} \]Adding these three, \(\displaystyle \mathrm{CO_2(g)} \) cancels (appears once on each side), \(\displaystyle \mathrm{2H_2O(l)} \) cancels (appears once on each side), and the oxygen on the right totals \(\displaystyle \tfrac{3}{2} + 1 + 1 = \tfrac{7}{2} \) mol while the oxygen on the left totals \(\displaystyle 1 + 1 = 2 \) mol, leaving \(\displaystyle \tfrac{7}{2} - 2 = \tfrac{3}{2} \) mol \(\displaystyle \mathrm{O_2}\) on the right side net. So the sum is exactly: \[\mathrm{C(graphite) + 2H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow CH_3OH(l)} \] which is the target reaction. Good — the same combination applies to the enthalpies.The step people skip: enthalpy is additive only when the equations are added exactly this way — reverse a reaction and you must reverse the sign of \(\displaystyle \Delta H\) too; scale a reaction by a factor and you must scale \(\displaystyle \Delta H\) by the same factor.\[\Delta_f H^0(\mathrm{CH_3OH,\, l}) = -\Delta_r H^0_{(i)} + \Delta_c H^0_{(ii)} + 2\,\Delta_f H^0_{(iii)} \]Substituting the values (all in kJ mol\(\displaystyle ^{-1}\), so no unit conversion is needed): \[\Delta_f H^0(\mathrm{CH_3OH,\, l}) = -(-726) + (-393) + 2(-286) \] \[= 726 - 393 - 572 \] \[= -239 \text{ kJ mol}^{-1} \]The negative sign shows the formation of methanol from graphite, hydrogen gas, and oxygen gas is exothermic, releasing $\displaystyle 239$ kJ of heat per mole of \(\displaystyle \mathrm{CH_3OH(l)}\) formed.Answer: \(\displaystyle \Delta_f H^0(\mathrm{CH_3OH,\, l}) = -239 \) kJ mol\(\displaystyle ^{-1}\)

Hess's Law lets you build an unknown enthalpy change from known ones — by inserting a path through the elements, even though the actual reaction never goes that way.The reaction to solve is the complete atomization of gaseous carbon tetrachloride: \[\text{CCl}_4(g) \rightarrow \text{C}(g) + 4\,\text{Cl}(g), \qquad \Delta H = \,? \]None of the given data describes this step directly. What's given are four separate thermochemical quantities, each for a different transformation: The trick people miss: you must build a path from \(\displaystyle \text{CCl}_4(g) \) back down to the elements, then back up to atoms — which means reversing the formation and vaporisation steps (flipping their signs).Route from \(\displaystyle \text{CCl}_4(g) \) to \(\displaystyle \text{C}(g) + 4\text{Cl}(g) \), going via the elements:Step $\displaystyle 1$ — condense the gas back to liquid (reverse of vaporisation): \[\text{CCl}_4(g) \rightarrow \text{CCl}_4(l), \qquad \Delta H_1 = -\Delta_{vap} H^{0} = -30.5 \text{ kJ mol}^{-1} \]Step $\displaystyle 2$ — decompose the liquid into its elements (reverse of formation): \[\text{CCl}_4(l) \rightarrow \text{C(graphite)} + 2\text{Cl}_2(g), \qquad \Delta H_2 = -\Delta_f H^{0} = -(-135.5) = +135.5 \text{ kJ mol}^{-1} \]Step $\displaystyle 3$ — atomise the carbon (used as given): \[\text{C(graphite)} \rightarrow \text{C}(g), \qquad \Delta H_3 = \Delta_a H^{0}(\text{C}) = +715.0 \text{ kJ mol}^{-1} \]Step $\displaystyle 4$ — atomise the chlorine. Here's the step that trips people up: the data is per mole of \(\displaystyle \text{Cl}_2 \), but Step $\displaystyle 2$ produced $\displaystyle 2$ mol of \(\displaystyle \text{Cl}_2 \), so the atomisation enthalpy must be doubled: \[2\text{Cl}_2(g) \rightarrow 4\text{Cl}(g), \qquad \Delta H_4 = 2 \times \Delta_a H^{0}(\text{Cl}_2) = 2 \times 242 = +484.0 \text{ kJ mol}^{-1} \]By Hess's Law, since Steps $\displaystyle 1$–$\displaystyle 4$ carry \(\displaystyle \text{CCl}_4(g) \) to exactly the same final state (\(\displaystyle \text{C}(g) + 4\text{Cl}(g) \)) as the direct reaction, their enthalpy changes simply add: \[\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4 \] \[\Delta H = (-30.5) + (135.5) + (715.0) + (484.0) \text{ kJ mol}^{-1} \] \[\Delta H = 1304.0 \text{ kJ mol}^{-1} \]This is the energy needed to break all four C–Cl bonds in one mole of \(\displaystyle \text{CCl}_4(g) \), turning it completely into isolated gaseous atoms.Bond enthalpy is an average, per-bond quantity — divide the total by the number of bonds broken, not by the number of atoms or moles of compound.Since the four C–Cl bonds are taken as identical, the mean bond enthalpy is: \[\text{Bond enthalpy (C–Cl)} = \frac{\Delta H}{4} = \frac{1304.0 \text{ kJ mol}^{-1}}{4} \] \[\text{Bond enthalpy (C–Cl)} = 326.0 \text{ kJ mol}^{-1} \]Answer: The enthalpy change for \(\displaystyle \text{CCl}_4(g) \rightarrow \text{C}(g) + 4\text{Cl}(g) \) is \(\displaystyle \Delta H = 1304.0 \text{ kJ mol}^{-1} \), and the bond enthalpy of C–Cl in \(\displaystyle \text{CCl}_4 \) is \(\displaystyle 326.0 \text{ kJ mol}^{-1} \).

An isolated system cannot exchange heat, work, or matter with its surroundings — so \(\displaystyle \Delta U = 0 \) only tells you energy is conserved inside the system; it says nothing yet about entropy, which is governed by a separate law.Step $\displaystyle 1$: What "isolated" forces on the first law.For an isolated system there is no heat flow and no work exchanged with the surroundings, so \(\displaystyle q = 0 \) and \(\displaystyle w = 0 \). The first law of thermodynamics, \[\Delta U = q + w, \] where \(\displaystyle \Delta U \) is the change in internal energy, \(\displaystyle q \) is heat absorbed by the system, and \(\displaystyle w \) is work done on the system, then gives \(\displaystyle \Delta U = 0 + 0 = 0 \). This is exactly the given condition — it is a restatement of energy conservation, not a statement about spontaneity.This is the step people trip on: seeing \(\displaystyle \Delta U = 0 \) and assuming "nothing changes," so \(\displaystyle \Delta S = 0 \) too. The first law (energy) and the second law (entropy) are independent statements — one being zero does not force the other to be zero.Step $\displaystyle 2$: Bring in the second law, which is what actually decides \(\displaystyle \Delta S \).The second law of thermodynamics says a process occurs spontaneously (on its own, without outside intervention) only if it increases the total entropy of the universe: \[\Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0 \quad \text{for a spontaneous process.} \] For a system that is isolated, there is no separate surroundings to exchange entropy with — the system is the entire universe as far as this process is concerned, so \(\displaystyle \Delta S_{\text{surroundings}} = 0 \) and the whole criterion collapses onto the system itself: \[\Delta S_{\text{system}} > 0 \quad \text{for a spontaneous change happening inside it.} \]Step $\displaystyle 3$: Check this against the definition of entropy.Entropy change along a reversible path is defined by \[dS = \frac{\delta q_{\text{rev}}}{T}, \] where \(\displaystyle \delta q_{\text{rev}} \) is heat added reversibly and \(\displaystyle T \) is the absolute temperature. If the change inside the isolated system could somehow happen reversibly, \(\displaystyle q = 0 \) would force \(\displaystyle dS = 0 \). But a change that occurs on its own — without any external agent driving it — is by definition irreversible, and the Clausius inequality for an irreversible step reads \[dS > \frac{\delta q}{T} = \frac{0}{T} = 0. \] So the real, spontaneous change proceeding inside the isolated system must have \(\displaystyle \Delta S > 0 \); only in the limiting, idealized case of a perfectly reversible process (no net spontaneous change at all) would \(\displaystyle \Delta S = 0 \).This is the same distinction as molarity vs. molality tripping people up elsewhere: two formulas that look like they should give the same sign of answer, but come from different physical premises. Here, "conserved" (first law, \(\displaystyle \Delta U \)) and "increasing" (second law, \(\displaystyle \Delta S \)) are not the same idea, and confusing them is the trap this question is built to catch.Answer: For an isolated system, \(\displaystyle \Delta S > 0 \) — entropy increases for the spontaneous process occurring within it (it equals zero only in the ideal limiting case of a reversible process).

A reaction becomes spontaneous when \(\displaystyle \Delta G \) turns negative, and \(\displaystyle \Delta G = \Delta H - T\Delta S \) is the equation that connects the three quantities you're given. Here \(\displaystyle \Delta H \) is the enthalpy change, \(\displaystyle \Delta S \) is the entropy change, and \(\displaystyle T \) is the absolute temperature in kelvin.For this reaction, \(\displaystyle \Delta H \) is large and positive (\(\displaystyle +400 \text{ kJ mol}^{-1} \)) — the reaction is strongly endothermic, so at low temperature the \(\displaystyle \Delta H \) term dominates and \(\displaystyle \Delta G \) is positive (non-spontaneous). The \(\displaystyle -T\Delta S \) term is what can pull \(\displaystyle \Delta G \) negative, and it only grows in magnitude as \(\displaystyle T \) increases. So there is a threshold temperature above which spontaneity kicks in.Finding that threshold: set \(\displaystyle \Delta G = 0 \). This is the boundary between non-spontaneous and spontaneous — the reaction is at equilibrium at exactly this temperature, and spontaneous ( \(\displaystyle \Delta G < 0\) ) for any temperature above it (since \(\displaystyle \Delta S > 0\) here).\[0 = \Delta H - T\Delta S \]Rearranging for \(\displaystyle T \):\[T = \frac{\Delta H}{\Delta S} \]A step people trip on: \(\displaystyle \Delta H \) is usually quoted in kJ but \(\displaystyle \Delta S \) is usually quoted in J — mixing those units without converting gives an answer wrong by a factor of 1000. Here both are already given in kilojoules, so no conversion is needed:\[\Delta H = 400 \text{ kJ mol}^{-1}, \qquad \Delta S = 0.2 \text{ kJ K}^{-1}\text{mol}^{-1} \]Substituting:\[T = \frac{400 \text{ kJ mol}^{-1}}{0.2 \text{ kJ K}^{-1}\text{mol}^{-1}} = 2000 \text{ K} \]The mol\(\displaystyle ^{-1}\) cancels top and bottom, and dividing kJ by (kJ K\(\displaystyle ^{-1}\)) leaves K — the unit comes out as a temperature, as it should.At \(\displaystyle T = 2000 \text{ K}\), \(\displaystyle \Delta G = 0\) exactly (the reaction sits at equilibrium). For any temperature above $\displaystyle 2000$ K, the \(\displaystyle -T\Delta S\) term outweighs \(\displaystyle \Delta H\), making \(\displaystyle \Delta G < 0\), so the reaction becomes spontaneous.Answer: The reaction becomes spontaneous above \(\displaystyle T = 2000 \text{ K} \) (with \(\displaystyle \Delta G = 0 \) exactly at $\displaystyle 2000$ K).

Forming a bond releases energy, and turning two free particles into one takes away freedom of motion — so here both \(\displaystyle \Delta H \) and \(\displaystyle \Delta S \) come out negative.Sign of \(\displaystyle \Delta H \) (enthalpy change):In \(\displaystyle 2\text{Cl(g)} \rightarrow \text{Cl}_2\text{(g)} \), two free chlorine atoms come together and form a single covalent bond, \(\displaystyle \text{Cl}-\text{Cl} \).Making a bond is always an energy-releasing process — you have to put in energy to break a bond (bond dissociation enthalpy is defined as a positive quantity for that reason), so the reverse process, forming the bond, gives that same amount of energy back out.Since the system releases energy to the surroundings, the reaction is exothermic, and by the sign convention (heat given out by the system is negative): \[\Delta H < 0 \]The step people get wrong: bond dissociation enthalpy (breaking a bond) is always quoted as positive; the moment a question is about forming that same bond, the sign flips to negative. Here the reaction is bond formation, not bond breaking, so \(\displaystyle \Delta H \) is negative even though the numerical value used to compute it (the Cl–Cl bond enthalpy) is tabulated as a positive number.Sign of \(\displaystyle \Delta S \) (entropy change):Entropy is a measure of the disorder, or the number of ways a system's particles can be arranged/move — more independent particles almost always means more disorder.On the left side there are $\displaystyle 2$ mol of separate Cl atoms, each free to move independently in the gas phase. On the right side these have combined into $\displaystyle 1$ mol of \(\displaystyle \text{Cl}_2 \) molecules — the same atoms now move together as a single, more constrained unit, and the total number of independently-moving gas particles has gone down ($\displaystyle 2$ mol of particles → $\displaystyle 1$ mol of particles).Fewer independent particles means fewer ways to arrange them, i.e., lower disorder. So the entropy of the system decreases going from reactants to products: \[\Delta S < 0 \]The step people get wrong: it is tempting to think "gas is involved, so entropy must increase" — but what matters is the change in the number of moles of gas (or more precisely, the change in the number of freely-moving particles), not merely the phase. Here the mole count of gas drops from $\displaystyle 2$ to $\displaystyle 1$, so entropy decreases, not increases.**Answer: \(\displaystyle \Delta H < 0 \) (negative, exothermic — a new Cl–Cl bond is formed) and \(\displaystyle \Delta S < 0 \) (negative — $\displaystyle 2$ mol of gaseous atoms combine into $\displaystyle 1$ mol of gaseous molecules, decreasing disorder).

\(\displaystyle \Delta U^\circ\) is not the same thing as \(\displaystyle \Delta H^\circ\) — for a gas-phase reaction where the mole count of gas changes, you must convert between them before you can use the Gibbs equation.The Gibbs equation needs \(\displaystyle \Delta H^\circ\), not \(\displaystyle \Delta U^\circ\): \[\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \] where \(\displaystyle \Delta H^\circ\) is the standard enthalpy change, \(\displaystyle T\) is the absolute temperature in kelvin, and \(\displaystyle \Delta S^\circ\) is the standard entropy change. The question gives you \(\displaystyle \Delta U^\circ\), so the first job is to get \(\displaystyle \Delta H^\circ\) from it.Step $\displaystyle 1$ — convert \(\displaystyle \Delta U^\circ\) to \(\displaystyle \Delta H^\circ\)For a reaction of ideal gases, \[\Delta H^\circ = \Delta U^\circ + \Delta n_g RT \] Here \(\displaystyle \Delta n_g\) is the change in the number of moles of gas (moles of gaseous products minus moles of gaseous reactants), \(\displaystyle R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}\) is the gas constant, and \(\displaystyle T\) is the temperature. This is the step it's easy to skip — treating \(\displaystyle \Delta U^\circ\) as if it were \(\displaystyle \Delta H^\circ\) — but whenever the number of gas moles changes, the two differ by a real, calculable amount.For \(\displaystyle 2A(g) + B(g) \rightarrow 2D(g)\): \[\Delta n_g = 2 - (2+1) = -1 \] The problem gives no temperature, so use the standard reference temperature implied by the "°" superscripts, \(\displaystyle T = 298\ \text{K}\).\[\Delta n_g RT = (-1)(8.314\ \text{J K}^{-1}\text{mol}^{-1})(298\ \text{K}) = -2477.6\ \text{J mol}^{-1} = -2.478\ \text{kJ mol}^{-1} \]\[\Delta H^\circ = \Delta U^\circ + \Delta n_g RT = (-10.5\ \text{kJ}) + (-2.478\ \text{kJ}) = -12.978\ \text{kJ} \]Step $\displaystyle 2$ — apply the Gibbs equationConvert \(\displaystyle \Delta S^\circ\) to kJ so the units match \(\displaystyle \Delta H^\circ\): \[\Delta S^\circ = -44.1\ \text{J K}^{-1} = -0.0441\ \text{kJ K}^{-1} \]Mixing J and kJ here is the other easy slip — always bring both terms to the same unit before subtracting.\[T\Delta S^\circ = (298\ \text{K})(-0.0441\ \text{kJ K}^{-1}) = -13.142\ \text{kJ} \]\[\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = (-12.978\ \text{kJ}) - (-13.142\ \text{kJ}) = -12.978\ \text{kJ} + 13.142\ \text{kJ} = 0.164\ \text{kJ} \]Step $\displaystyle 3$ — read the sign\(\displaystyle \Delta G^\circ\) has come out positive. A positive \(\displaystyle \Delta G^\circ\) means the reaction is not spontaneous in the forward direction under standard conditions at $\displaystyle 298$ K — even though \(\displaystyle \Delta U^\circ\) itself was negative. The internal-energy change alone does not tell you about spontaneity; only \(\displaystyle \Delta G^\circ\), which weighs in the entropy term as well, does.Answer: \(\displaystyle \Delta G^\circ \approx +0.164\ \text{kJ}\) (i.e. about \(\displaystyle +164\ \text{J}\)); since \(\displaystyle \Delta G^\circ > 0\), the reaction is not spontaneous under standard conditions at $\displaystyle 298$ K.

The link between the equilibrium constant and \(\displaystyle \Delta G^0\) is \(\displaystyle \Delta G^0 = -RT\ln K\), and this is exact — no approximation.Here \(\displaystyle \Delta G^0\) is the standard Gibbs energy change of the reaction, \(\displaystyle R\) is the gas constant, \(\displaystyle T\) is the absolute temperature in kelvin, and \(\displaystyle K\) is the equilibrium constant. This equation is what tells you whether a reaction favours products (\(\displaystyle K>1\), \(\displaystyle \Delta G^0<0\)) or reactants (\(\displaystyle K<1\), \(\displaystyle \Delta G^0>0\)) at equilibrium.A step people rush past: \(\displaystyle T\) must be in kelvin, not degrees Celsius — the formula only works with absolute temperature, and here it is already given as $\displaystyle 300$ K.Given: \[K = 10, \quad R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}, \quad T = 300\ \text{K} \]Substitute directly into \(\displaystyle \Delta G^0 = -RT\ln K\): \[\Delta G^0 = -(8.314\ \text{J K}^{-1}\text{mol}^{-1})(300\ \text{K})\,\ln(10) \]A convenient rewrite people forget: \(\displaystyle \ln x = 2.303 \log_{10} x\), so \(\displaystyle \ln(10) = 2.303 \times \log_{10}(10) = 2.303 \times 1 = 2.303\). This lets the whole calculation run on base-$\displaystyle 10$ logs.\[\Delta G^0 = -(8.314\ \text{J K}^{-1}\text{mol}^{-1})(300\ \text{K})(2.303) \]Multiply the numbers in stages, keeping units attached: \[8.314\ \text{J K}^{-1}\text{mol}^{-1} \times 300\ \text{K} = 2494.2\ \text{J mol}^{-1} \] \[2494.2\ \text{J mol}^{-1} \times 2.303 = 5744.1\ \text{J mol}^{-1} \]So: \[\Delta G^0 = -5744.1\ \text{J mol}^{-1} \]The data (\(\displaystyle K=10\), \(\displaystyle T = 300\) K) supports three significant figures, so this rounds to \(\displaystyle -5.74 \times 10^{3}\ \text{J mol}^{-1}\), i.e. \(\displaystyle -5.74\ \text{kJ mol}^{-1}\).The negative sign is not incidental — it says something physical: because \(\displaystyle K > 1\), the reaction is product-favoured at equilibrium, and \(\displaystyle \Delta G^0\) must come out negative. A positive \(\displaystyle \Delta G^0\) would instead signal \(\displaystyle K < 1\) (reactant-favoured), so the sign is a built-in check on the arithmetic, not just a label.Answer: \(\displaystyle \Delta G^0 = -5.74\ \text{kJ mol}^{-1}\) (\(\displaystyle -5744\ \text{J mol}^{-1}\)).

The sign of \(\displaystyle \Delta_fH^\circ \) tells you where a compound sits on the energy scale relative to its own elements — positive means "uphill," and uphill compounds are thermodynamically unstable with respect to falling back to those elements.Balancing each equation so that exactly one mole of product is formed (that is what makes these enthalpies of formation, \(\displaystyle \Delta_fH^\circ \), and not just reaction enthalpies for an arbitrary equation) fixes the fractional coefficients:\[\text{(1)}\quad \tfrac{1}{2}\text{N}_2(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{NO}(g); \qquad \Delta_rH_1^\circ = +90\ \text{kJ mol}^{-1} \]\[\text{(2)}\quad \text{NO}(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{NO}_2(g); \qquad \Delta_rH_2^\circ = -74\ \text{kJ mol}^{-1} \]The easy slip here is dropping the \(\displaystyle \tfrac{1}{2} \) in front of \(\displaystyle \text{N}_2 \) and \(\displaystyle \text{O}_2 \) — an enthalpy of formation is always defined per mole of the compound formed, and one mole of NO only needs half a mole of each diatomic element.Step $\displaystyle 1$ — read reaction ($\displaystyle 1$) on its own. Reaction ($\displaystyle 1$) is literally the formation reaction of NO from its elements in their standard states, so \[\Delta_fH^\circ(\text{NO}) = \Delta_rH_1^\circ = +90\ \text{kJ mol}^{-1}. \] This is strongly positive: making NO out of \(\displaystyle \text{N}_2 \) and \(\displaystyle \text{O}_2 \) costs $\displaystyle 90$ kJ for every mole formed. Equivalently, the reverse process — NO(g) breaking back down into \(\displaystyle \text{N}_2(g) \) and \(\displaystyle \text{O}_2(g) \) — releases $\displaystyle 90$ kJ mol⁻¹, an exothermic, energetically "downhill" path. A compound that can fall downhill in energy back to its elements like this is, by definition, thermodynamically unstable relative to those elements.Step $\displaystyle 2$ — use Hess's law to see what happens on the way to \(\displaystyle \text{NO}_2 \). Adding equation ($\displaystyle 1$) and equation ($\displaystyle 2$), the intermediate NO(g) cancels (one mole is produced in ($\displaystyle 1$) and consumed in ($\displaystyle 2$)):\[\tfrac{1}{2}\text{N}_2(g) + \tfrac{1}{2}\text{O}_2(g) + \text{NO}(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{NO}(g) + \text{NO}_2(g) \] \[\Longrightarrow\ \tfrac{1}{2}\text{N}_2(g) + \text{O}_2(g) \rightarrow \text{NO}_2(g) \]Because enthalpy is a state function, the enthalpy of this net reaction is just the sum of the two steps (Hess's Law: \(\displaystyle \Delta_fH^\circ = \Delta_rH_1^\circ + \Delta_rH_2^\circ \), where each \(\displaystyle \Delta_rH^\circ \) is the heat absorbed or released by one step of the path):\[\Delta_fH^\circ(\text{NO}_2) = (+90\ \text{kJ mol}^{-1}) + (-74\ \text{kJ mol}^{-1}) = +16\ \text{kJ mol}^{-1} \]Step $\displaystyle 3$ — compare the two compounds. Both \(\displaystyle \Delta_fH^\circ \) values are positive, so both NO(g) and \(\displaystyle \text{NO}_2(g) \) sit above \(\displaystyle \text{N}_2(g) + \text{O}_2(g) \) on the energy scale — neither is stable relative to the free elements. But +$\displaystyle 90$ kJ mol⁻¹ is a much steeper climb than +$\displaystyle 16$ kJ mol⁻¹, so NO(g) is the less stable of the two: it sits higher up the hill, and rolling it further downhill by oxidising it to \(\displaystyle \text{NO}_2(g) \) releases a large $\displaystyle 74$ kJ mol⁻¹, which is exactly why reaction ($\displaystyle 2$) runs in the exothermic direction as written.Answer: \(\displaystyle \Delta_fH^\circ(\text{NO}) = +90\ \text{kJ mol}^{-1} \) is large and positive, so NO(g) is thermodynamically unstable with respect to decomposition into \(\displaystyle \text{N}_2(g) \) and \(\displaystyle \text{O}_2(g) \) (that reverse step is exothermic by $\displaystyle 90$ kJ mol⁻¹). Combining the two given steps by Hess's Law gives \(\displaystyle \Delta_fH^\circ(\text{NO}_2) = 90 + (-74) = +16\ \text{kJ mol}^{-1} \); since this is a much smaller positive value than for NO, \(\displaystyle \text{NO}_2(g) \) is comparatively more stable than NO(g), though still less stable than the free elements \(\displaystyle \text{N}_2(g) \) and \(\displaystyle \text{O}_2(g) \).

Heat leaving the system enters the surroundings — that's what changes the surroundings' entropy, not the reaction happening inside the system.When a reaction runs at constant temperature and pressure, the heat it releases (or absorbs) is transferred to the surroundings so slowly and evenly that the surroundings can be treated as absorbing it reversibly. That gives a formula that connects the system's enthalpy change directly to the surroundings' entropy change:\[\Delta S_{\text{surr}} = \dfrac{-\Delta H_{\text{system}}}{T} \]Here \(\displaystyle \Delta H_{\text{system}}\) is the enthalpy change of the reaction happening in the system (what NCERT calls \(\displaystyle \Delta_f H^0\) for a formation reaction), \(\displaystyle T\) is the absolute temperature in kelvin, and the minus sign is doing real work: heat leaving the system (\(\displaystyle \Delta H_{\text{system}} < 0\), exothermic) is heat entering the surroundings, so it makes \(\displaystyle \Delta S_{\text{surr}}\) positive.Set up the numbers. The reaction is the formation of liquid water, \[\mathrm{H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow H_2O(l)}, \qquad \Delta_f H^0 = -286\ \text{kJ mol}^{-1} \] "Standard conditions" fixes the temperature at \(\displaystyle T = 298\ \text{K}\) (this is the temperature standard thermodynamic data is tabulated at, unless a different one is stated).Before substituting, convert kilojoules to joules — mixing kJ into a formula whose answer you want in J K⁻¹ mol⁻¹ is a common slip: \[\Delta_f H^0 = -286\ \text{kJ mol}^{-1} = -286 \times 10^{3}\ \text{J mol}^{-1} = -2.86\times10^{5}\ \text{J mol}^{-1} \]Substitute. \[\Delta S_{\text{surr}} = \dfrac{-\left(-2.86\times10^{5}\ \text{J mol}^{-1}\right)}{298\ \text{K}} = \dfrac{2.86\times10^{5}\ \text{J mol}^{-1}}{298\ \text{K}} \]The two negative signs (the minus in the formula, and the minus already inside \(\displaystyle \Delta_f H^0\)) cancel — that's expected, since releasing heat from an exothermic reaction should raise the surroundings' entropy, not lower it.\[\Delta S_{\text{surr}} = 959.73\ \text{J K}^{-1}\text{mol}^{-1} \]The data ($\displaystyle 286$ kJ mol⁻¹, three significant figures) justifies rounding the final value to three significant figures:\[\Delta S_{\text{surr}} \approx 960\ \text{J K}^{-1}\text{mol}^{-1} \]Answer: \(\displaystyle \Delta S_{\text{surroundings}} = \dfrac{2.86\times10^{5}\ \text{J mol}^{-1}}{298\ \text{K}} \approx 960\ \text{J K}^{-1}\text{mol}^{-1}\) (positive, because the exothermic reaction releases heat into the surroundings).