Equilibrium MCQs with Answers – Part 1 (NEET Chemistry)

Explanation: A reversible chemical reaction can proceed in both the forward and reverse directions. Initially, the two opposing reaction rates are generally unequal, causing the composition of the mixture to change. As the reaction proceeds, the concentrations of reactants and products change until a state of dynamic equilibrium is established. At equilibrium, the forward and reverse reaction rates become equal. Consequently, the rate of formation of each reacting species is balanced by its rate of consumption. This produces constant macroscopic concentrations even though chemical transformations continue at the molecular level. Equal forward and reverse rates do not imply that reactant and product concentrations are numerically equal. Depending on the reaction and conditions, the equilibrium mixture may contain predominantly reactants, predominantly products, or appreciable amounts of both. Complete conversion is also not a necessary condition for equilibrium. The decisive characteristic is the absence of net compositional change while the opposing reactions continue.

Explanation: The equilibrium condition for a reversible chemical reaction is equality of its forward and reverse reaction rates. It is not equality of the concentrations of the substances participating in the reaction. The given mixture contains four times as much \(A\) as \(B\) in terms of molar concentration. This difference does not prevent the opposing reactions from balancing one another. Reaction rates depend on the nature of the reaction and its conditions, not simply on which species has the greater concentration. At equilibrium, each forward transformation is statistically balanced by the corresponding reverse transformation. Therefore, there is no net conversion of \(A\) into \(B\), even though individual molecules continue to react. The concentrations remain fixed at their equilibrium values as long as the conditions are maintained. If the forward rate were greater than the reverse rate, \(A\) would undergo net consumption. If the reverse rate were greater, \(B\) would undergo net consumption. Neither situation is consistent with the stated equilibrium composition. Thus, unequal equilibrium concentrations are entirely compatible with equal opposing reaction rates.

Explanation: Constant concentration is an important characteristic of chemical equilibrium, but it must be interpreted together with the mechanism maintaining that concentration. In System P, the vessel is closed and the opposing chemical transformations balance one another. No external addition or removal of reacting substances is required to maintain the composition. The equality of forward and reverse reaction rates therefore establishes dynamic chemical equilibrium. System Q is fundamentally different because matter continuously enters and leaves the reaction vessel. Its concentrations can remain unchanged even when the forward and reverse chemical reaction rates are unequal. The incoming material can compensate for chemical consumption, while product withdrawal can compensate for chemical formation. Such a condition is a steady state maintained by external material flows. A steady state does not automatically satisfy the conditions for thermodynamic equilibrium. If the external supply and withdrawal are stopped, the composition of System Q may begin changing. The key distinction is whether macroscopic constancy results from balanced opposing chemical transformations or from externally maintained material flows. Therefore, equilibrium is established for P, but the supplied information does not establish equilibrium for Q.

Explanation: \(\textbf{Identify the conservation relationship:}\) The balanced reaction shows that two units of \(A\) are converted into one unit of \(B\). Consequently, the quantity \([A]+2[B]\) remains conserved in a closed, constant-volume vessel. \[ [A]+2[B]=\text{constant} \] \(\textbf{Compare the initial preparations:}\) For vessel P, \[ [A]_0+2[B]_0=1.00+2(0)=1.00\,\mathrm{M} \] For vessel Q, \[ [A]_0+2[B]_0=0+2(0.50)=1.00\,\mathrm{M} \] Thus, both vessels have the same conserved amount of the reacting material per unit volume. They also have the same temperature and undergo the same reversible chemical transformation. \(\textbf{Check the established equilibrium in P:}\) \[ [A]_{\mathrm{eq}}+2[B]_{\mathrm{eq}} =0.40+2(0.30) \] \[ =1.00\,\mathrm{M} \] The stated equilibrium composition satisfies the required conservation relationship. \(\textbf{Reason from the opposite starting direction:}\) In P, the reaction initially proceeds predominantly towards the formation of \(B\). In Q, the initial composition favours the reverse conversion of \(B\) into \(A\). However, the equilibrium state under these identical conditions is determined by the same reversible system and conserved composition, not by the direction from which equilibrium is approached. The two vessels can therefore approach the same equilibrium composition from opposite directions. \(\textbf{Final answer:}\) \[ \boxed{[A]_{\mathrm{eq}}=0.40\,\mathrm{M},\quad [B]_{\mathrm{eq}}=0.30\,\mathrm{M}} \] The different initial compositions affect the paths towards equilibrium, but not the final equilibrium composition in this case.

Explanation: The Assertion describes the central characteristic of dynamic chemical equilibrium. A reversible reaction does not necessarily cease when the equilibrium state is reached. Instead, reactant molecules continue forming products while product molecules continue regenerating reactants. The crucial difference between an ordinary reacting mixture and an equilibrium mixture is the balance of these opposing processes. The Reason correctly identifies equality of the forward and reverse reaction rates as the condition for this balance. For a properly defined reaction rate, the net rate of reaction is the forward rate minus the reverse rate. At equilibrium, this difference becomes zero. Therefore, there is no net change in the amounts of reactants or products due to the reaction. This explains why measurable concentrations can remain constant despite continuing molecular activity. The Reason is not merely another correct statement about equilibrium; it identifies the mechanism responsible for the phenomenon described in the Assertion. It also resolves the common misconception that unchanged concentrations imply the complete cessation of chemical reactions. Hence, both statements are true, and the Reason correctly explains the Assertion.

Explanation: \(\textbf{Determine the initially possible reaction:}\) The vessel initially contains only \(B\), so the reverse transformation \(B\to A\) can begin immediately. There is initially no \(A\) available for the forward transformation \(A\to B\). Therefore, the forward reaction rate is initially zero, whereas the reverse rate is positive. \(\textbf{Follow the composition changes:}\) As the reverse reaction proceeds, \(A\) begins to accumulate. The newly formed \(A\) can undergo the forward reaction, causing the forward rate to become positive. Meanwhile, the concentration of \(B\) changes as the system approaches equilibrium. Under the usual reversible-reaction behaviour, the initially dominant reverse rate decreases while the forward rate increases. \(\textbf{Identify the equilibrium condition:}\) The two rates eventually become equal. They do not have to become zero because equilibrium is dynamic. A graph showing equal positive limiting rates is therefore consistent with the established equilibrium state. \(\textbf{Check the direction of approach:}\) The forward and reverse rate curves need not have equal initial values. Their initial relative magnitudes depend on which reacting species are present. Starting with pure \(B\) reverses the usual initial-rate pattern observed when starting with pure \(A\). The correctly described graph must reflect this initial condition as well as the final equality of rates. Hence, option B is correct.

Explanation: \(\textbf{Examine the reaction stoichiometry:}\) The reaction converts one mole of gaseous \(A\) into one mole of gaseous \(B\). Therefore, the forward transformation does not change the total number of gaseous moles. The reverse transformation also conserves the total number of gaseous moles. \(\textbf{Relate pressure to the gaseous state:}\) For an ideal gas mixture, \[ PV=n_{\mathrm{total}}RT \] The vessel is rigid, so \(V\) remains constant. Its temperature is also maintained constant. Because the reaction preserves \(n_{\mathrm{total}}\), the total pressure remains unchanged throughout the transformation. \(\textbf{Separate total pressure from composition:}\) The constancy of total pressure does not mean that the individual amounts of \(A\) and \(B\) are fixed. As \(A\) is converted into \(B\), the partial pressure of \(A\) can decrease while that of \(B\) increases. Their sum can remain equal to the original total pressure. Thus, a constant total pressure is compatible with a changing chemical composition. \(\textbf{Apply the equilibrium criterion:}\) Chemical equilibrium requires the net reaction rate to become zero because the forward and reverse rates are equal. The given analysis explicitly states that the mole fractions are changing initially. Consequently, the system has not yet attained chemical equilibrium during that interval. The constant total pressure is a consequence of gaseous stoichiometry, not proof of equilibrium. Option D correctly distinguishes conservation of total gaseous moles from the dynamic equilibrium condition.

Explanation: The equality of concentrations at one instant is not the defining criterion for chemical equilibrium. The relevant condition is equality of the forward and reverse reaction rates. Here, the forward rate is greater than the reverse rate. Therefore, more \(A\) is being converted into \(B\) than \(B\) is being converted back into \(A\). The reaction mixture consequently undergoes a net change towards greater formation of \(B\). Although \([A]=[B]\) at \(t_1\), that equality does not remain preserved while the net forward conversion continues. At equilibrium, the net rate of conversion would have to vanish. The measured rates explicitly exclude this possibility. The fact that both rates are nonzero is consistent with a reversible reaction, but it is insufficient to establish equilibrium. A dynamic equilibrium requires both opposing processes to occur at equal rates, not merely to occur simultaneously. The example also shows that concentration curves can intersect before an equilibrium state is attained. Hence, the mixture is not at equilibrium and the net reaction proceeds towards \(B\).

Explanation: \(\textbf{Interpret the initial observations:}\) At P, the forward reaction rate exceeds the reverse reaction rate. The mixture therefore undergoes net conversion of \(A\) into \(B\). However, the reverse reaction is already occurring because its measured rate is positive. At Q, the difference between the two rates is smaller than at P. Nevertheless, the forward reaction is still faster, so the mixture has not yet attained equilibrium. \(\textbf{Identify the balanced-rate state:}\) At R, the forward and reverse reaction rates become equal. At S, their equality is maintained under the same external conditions. The additional observation that the concentrations remain constant between R and S confirms that the mixture has reached a stable equilibrium composition. \(\textbf{Distinguish net rate from individual rates:}\) The net reaction rate becomes zero at equilibrium, but neither opposing rate is zero. The equality \(r_{\mathrm{forward}}=r_{\mathrm{reverse}}\) represents a balance of continuing chemical transformations. It does not mean that the reaction has become irreversible. An irreversible reaction would not be described by two continuing, measurable opposing transformations. The observations at R and S therefore support dynamic equilibrium. The decisive evidence is sustained equality of the opposing rates together with constant composition, not merely the simultaneous existence of forward and reverse reactions.

Explanation: Chemical equilibrium is a dynamic state in which the forward and reverse reactions continue simultaneously. The equality of these rates prevents a net change in the overall concentrations of the reacting substances. However, individual molecules are not required to remain permanently in their original chemical forms. Introducing a chemically equivalent isotopic label makes it possible to follow the movement of selected atoms through the reversible process. When labelled \(A^*\) undergoes the forward transformation, labelled \(B^*\) can be produced. At the same time, the reverse transformation continues converting \(B\)-type molecules into \(A\)-type molecules. The total amount of \(B\) need not increase because its rate of formation is balanced by its rate of consumption. The appearance of labelled product therefore reveals molecular exchange that cannot be detected by ordinary concentration measurements alone. The isotopic label acts as a tracer rather than as a new driving force for the reaction under the stated conditions. An equilibrium mixture can consequently exhibit changes in the identities of individual molecules while maintaining constant macroscopic composition. This distinction is important because static equilibrium would incorrectly imply that molecular transformations had ceased. The appearance of \(B^*\) with unchanged total concentrations demonstrates the dynamic nature of the equilibrium state.

Explanation: \(\textbf{Identify the reaction and supplied rates:}\) \[ 2A(g)\rightleftharpoons B(g) \] \[ r_f=5\,\mathrm{mmol\,L^{-1}\,min^{-1}} \] \[ r_r=3\,\mathrm{mmol\,L^{-1}\,min^{-1}} \] The supplied rates refer to reaction events, so stoichiometric coefficients must be considered when calculating the disappearance of a particular species. \(\textbf{Determine the net reaction rate:}\) The forward reaction consumes \(A\), whereas the reverse reaction regenerates it. Therefore, \[ r_{\mathrm{net}}=r_f-r_r \] \[ r_{\mathrm{net}}=5-3 \] \[ r_{\mathrm{net}}=2\,\mathrm{mmol\,L^{-1}\,min^{-1}} \] The positive net rate indicates that the forward direction is dominant. \(\textbf{Apply the stoichiometric coefficient of A:}\) Two units of \(A\) disappear for each net forward reaction event. Hence, \[ -\frac{d[A]}{dt}=2r_{\mathrm{net}} \] Substituting the net reaction rate, \[ -\frac{d[A]}{dt}=2(2) \] \[ -\frac{d[A]}{dt}=4\,\mathrm{mmol\,L^{-1}\,min^{-1}} \] \(\textbf{Chemical consistency check:}\) The net rate of formation of \(B\) is \(2\,\mathrm{mmol\,L^{-1}\,min^{-1}}\). The disappearance rate of \(A\) must be twice that value because the balanced equation requires two units of \(A\) for one unit of \(B\). The rates of the opposing reactions are unequal, so the system is not at equilibrium at the stated instant. The answer cannot be obtained merely by subtracting the reaction-event rates and treating that difference as the disappearance rate of \(A\). \(\textbf{Final answer:}\) \[ \boxed{-\frac{d[A]}{dt}=4\,\mathrm{mmol\,L^{-1}\,min^{-1}}} \]

Explanation: \(\textbf{Interpret the gaseous stoichiometry:}\) The forward reaction converts two moles of \(A\) into one mole of \(B\). Consequently, the total number of gaseous moles decreases as the net reaction proceeds in the forward direction. \[ 2A(g)\rightleftharpoons B(g) \] The decrease in gaseous particle number provides the first explanation for the observed pressure change. \(\textbf{Connect pressure with the vessel conditions:}\) The ideal gas equation is \[ PV=n_{\mathrm{total}}RT \] Temperature and volume are both constant. Therefore, \[ P\propto n_{\mathrm{total}} \] The decreasing total pressure is consistent with the net formation of \(B\), which reduces the total number of gaseous moles. \(\textbf{Interpret the approach to the plateau:}\) Initially, only \(A\) is present, so the net forward transformation begins as \(B\) forms. As \(B\) accumulates, the reverse transformation also becomes increasingly significant. The difference between the opposing reaction rates decreases during the usual approach to equilibrium. When the rates become equal, the net change in the reaction composition becomes zero. Consequently, the total number of gaseous moles also becomes constant. \(\textbf{Check the equilibrium interpretation:}\) The pressure plateau is accompanied by constant individual species concentrations, as specified in the question. These observations are consistent with the attainment of dynamic equilibrium. They do not imply that molecular motion or chemical transformations have stopped. They also do not require complete disappearance of \(A\) or equal concentrations of \(A\) and \(B\). The decreasing pressure is therefore explained by gaseous stoichiometry, while the eventual plateau reflects the absence of further net chemical conversion. Hence, option A correctly connects both features of the graph.

Explanation: At approximately atmospheric pressure, \(273\,\mathrm{K}\) is the equilibrium melting temperature of pure ice. At this temperature and pressure, ice and liquid water can coexist in equilibrium. The absence of measurable changes in their masses does not mean that molecular exchange has stopped. Molecules at the ice-water boundary continually move between the ordered solid structure and the liquid phase. Some ice molecules enter the liquid phase through melting. Simultaneously, liquid water molecules become incorporated into the solid through freezing. At equilibrium, these opposing phase-change processes occur at equal rates. Therefore, the amount of ice lost through melting is balanced by the amount gained through freezing. The same balance maintains the mass of the liquid phase. The thermal insulation prevents external heat exchange that could disturb the stated equilibrium conditions. The temperature remains fixed at the coexistence temperature while both phases remain present under the specified pressure. Thus, unchanged phase masses result from balanced molecular exchange rather than cessation of melting and freezing.

Explanation: \(\textbf{Identify the initial phase-equilibrium condition:}\) Ice and liquid water initially coexist at their melting temperature under atmospheric pressure. The supplied energy can therefore be used to convert part of the ice into liquid water. While both phases remain present at the stated pressure, the temperature remains at the equilibrium melting point. \(\textbf{Write the supplied quantities:}\) \[ m_{\mathrm{ice,initial}}=150\,\mathrm{g} \] \[ m_{\mathrm{water,initial}}=200\,\mathrm{g} \] \[ Q=33.4\,\mathrm{kJ}=33400\,\mathrm{J} \] \[ L_f=334\,\mathrm{J\,g^{-1}} \] \(\textbf{Apply the latent-heat relation:}\) \[ Q=m_{\mathrm{melted}}L_f \] Rearranging, \[ m_{\mathrm{melted}}=\frac{Q}{L_f} \] Substitution gives \[ m_{\mathrm{melted}}=\frac{33400}{334} \] \[ m_{\mathrm{melted}}=100\,\mathrm{g} \] \(\textbf{Determine the remaining ice:}\) The original ice mass is \(150\,\mathrm{g}\). Of this, \(100\,\mathrm{g}\) melts. \[ m_{\mathrm{ice,final}}=150-100 \] \[ m_{\mathrm{ice,final}}=50\,\mathrm{g} \] \(\textbf{Determine the final liquid mass:}\) The melted ice becomes liquid water and joins the initially present water. \[ m_{\mathrm{water,final}}=200+100 \] \[ m_{\mathrm{water,final}}=300\,\mathrm{g} \] \(\textbf{Verify the thermal state:}\) There is still \(50\,\mathrm{g}\) of unmelted ice. Therefore, both solid and liquid phases remain present. At the specified atmospheric pressure, their equilibrium coexistence temperature is \(0^\circ\mathrm{C}\). The supplied energy has been consumed by melting and does not raise the final equilibrium temperature. The total mass remains \(350\,\mathrm{g}\), confirming mass conservation. \(\textbf{Final answer:}\) \[ \boxed{m_{\mathrm{ice}}=50\,\mathrm{g},\quad m_{\mathrm{water}}=300\,\mathrm{g},\quad T=0^\circ\mathrm{C}} \]

Explanation: \(\textbf{Identify the relevant phase-volume relationship:}\) Ice has a more open solid structure than liquid water under ordinary conditions. Consequently, ice is less dense and has a larger molar volume than liquid water. Melting ice therefore produces a decrease in volume. \(\textbf{Predict the effect of pressure:}\) Increasing pressure favours the phase occupying the smaller volume. For the ice-water system, this is the liquid phase. Therefore, increasing pressure promotes melting relative to freezing at the original coexistence temperature. \(\textbf{Infer the new melting condition:}\) Because liquid water becomes relatively more favourable at higher pressure, equilibrium coexistence is restored at a slightly lower temperature. The melting temperature of ice consequently decreases with increasing pressure over the ordinary range considered. This behaviour differs from that of many substances whose solid phase is denser than their liquid phase. For such ordinary substances, increasing pressure generally favours the solid and raises the melting temperature. \(\textbf{Distinguish phase preference from reaction rate:}\) The direction of the equilibrium displacement is determined by the relative phase volumes. It is not necessary for the two phases to have equal masses. Nor does increasing pressure simply stop molecular exchange between ice and water. If both phases remain available at the new coexistence conditions, melting and freezing again occur at equal rates. The unusual pressure dependence of ice melting follows from the fact that liquid water occupies less volume than ice. Hence, greater pressure favours the liquid phase and lowers the melting temperature.

Explanation: \(\textbf{Identify the initial equilibrium:}\) Liquid water and its vapour initially coexist at a fixed temperature. At equilibrium, the rate of evaporation equals the rate of condensation. The vapour exerts the equilibrium vapour pressure characteristic of water at that temperature. \(\textbf{Analyse the immediate volume change:}\) When the piston is suddenly withdrawn, the vapour occupies a larger volume. The number of vapour molecules does not increase instantaneously by the amount required to maintain the original vapour density. Therefore, the vapour pressure initially decreases. The system has temporarily moved away from liquid-vapour equilibrium. \(\textbf{Determine the response of opposing processes:}\) At the unchanged temperature, liquid water molecules continue escaping from the liquid surface. The reduced vapour density produces fewer vapour molecules returning to the liquid per unit time. Consequently, evaporation initially exceeds condensation. There is a net transfer of water molecules from the liquid phase into the vapour phase. \(\textbf{Infer the new equilibrium state:}\) As additional vapour accumulates, its pressure rises. The condensation rate increases until it again balances the evaporation rate. At the original temperature, the equilibrium vapour pressure of pure water is the same as before, provided liquid water remains present. The final vapour pressure therefore returns to its initial equilibrium value. The final amount of vapour is larger because the available vapour volume is larger, but the equilibrium pressure is unchanged. The reasoning requires distinguishing the immediate disturbance from the final equilibrium response. It also shows why equilibrium vapour pressure depends primarily on temperature rather than on the available volume when both liquid and vapour coexist. Hence, option C correctly describes both the transient change and the final state.

Explanation: \(\textbf{Examine the two reversible transformations:}\) The first equilibrium pair involves interconversion between \(A\) and \(B\). The second involves interconversion between \(B\) and \(C\). Because \(B\) participates in both reactions, its concentration change depends on contributions from both reaction pairs. \(\textbf{Net conversion in the first reaction:}\) \[ r_{A\to B,\mathrm{net}}=7-3 \] \[ r_{A\to B,\mathrm{net}}=4\,\mathrm{mmol\,L^{-1}\,min^{-1}} \] Thus, the first reaction produces \(B\) at a net rate of \(4\,\mathrm{mmol\,L^{-1}\,min^{-1}}\). \(\textbf{Net conversion in the second reaction:}\) \[ r_{B\to C,\mathrm{net}}=6-2 \] \[ r_{B\to C,\mathrm{net}}=4\,\mathrm{mmol\,L^{-1}\,min^{-1}} \] Thus, the second reaction consumes \(B\) at exactly the same net rate. \(\textbf{Determine the concentration change of B:}\) \[ \frac{d[B]}{dt}=(7-3)-(6-2) \] \[ \frac{d[B]}{dt}=4-4 \] \[ \frac{d[B]}{dt}=0 \] Although the instantaneous net concentration change of \(B\) is zero, the other species are still changing. \[ \frac{d[A]}{dt}=-4\,\mathrm{mmol\,L^{-1}\,min^{-1}} \] \[ \frac{d[C]}{dt}=+4\,\mathrm{mmol\,L^{-1}\,min^{-1}} \] \(\textbf{Apply the equilibrium criterion:}\) Chemical equilibrium requires the opposing reactions within each reversible pair to balance. Here, the first pair has rates \(7\) and \(3\), while the second has rates \(6\) and \(2\). Neither pair has equal opposing rates. The stationary concentration of \(B\) results from cancellation between two different net chemical conversions. It does not establish equilibrium for the complete reacting system. The example demonstrates why observing one unchanged species concentration is insufficient to establish chemical equilibrium. Hence, option B is correct.

Explanation: \(\textbf{Interpret the forward transformation:}\) The balanced reaction shows that two moles of \(A\) are consumed whenever three moles of \(B\) are formed. The observed forward consumption of \(A\) is \(0.20\,\mathrm{mol}\). Therefore, the amount of \(B\) formed through the forward process is obtained from the stoichiometric ratio. \[ n_{B,\mathrm{formed}}=0.20\times\frac{3}{2} \] \[ n_{B,\mathrm{formed}}=0.30\,\mathrm{mol} \] \(\textbf{Use the dynamic-equilibrium condition:}\) At equilibrium, the forward and reverse rates of reaction are equal. Hence, over the specified observation interval, the average extents of the forward and reverse transformations are equal. The reverse reaction must therefore consume the same amount of \(B\) as the forward reaction produces. \[ n_{B,\mathrm{reverse\ consumed}}=0.30\,\mathrm{mol} \] \(\textbf{Calculate the net formation of B:}\) \[ \Delta n_B=n_{B,\mathrm{forward\ formed}}-n_{B,\mathrm{reverse\ consumed}} \] \[ \Delta n_B=0.30-0.30 \] \[ \Delta n_B=0\,\mathrm{mol} \] \(\textbf{Interpret the result:}\) The zero net change does not imply that no molecules of \(B\) have been produced or consumed. There is continuous molecular turnover through both directions of the reversible reaction. The reaction stoichiometry determines the quantities of individual chemical species participating in that turnover. The dynamic-equilibrium condition determines why the opposing transformations cancel macroscopically. It would be incorrect to equate the \(0.20\,\mathrm{mol}\) of \(A\) consumed with the amount of \(B\) consumed without accounting for the coefficients. \(\textbf{Final answer:}\) \[ \boxed{n_{B,\mathrm{reverse\ consumed}}=0.30\,\mathrm{mol},\quad \Delta n_B=0\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify the initial thermal conditions:}\) The water is initially at \(0^\circ\mathrm{C}\), while the introduced ice is colder. The ice must gain heat to reach the equilibrium melting temperature. Because the water is already at \(0^\circ\mathrm{C}\), it cannot supply this heat merely by cooling while remaining liquid at atmospheric pressure. A small amount of liquid water must freeze, releasing latent heat. \(\textbf{Determine the heat needed to warm the ice:}\) \[ Q=mc_{\mathrm{ice}}\Delta T \] Here, \[ m=20\,\mathrm{g} \] \[ c_{\mathrm{ice}}=2.1\,\mathrm{J\,g^{-1}\,K^{-1}} \] \[ \Delta T=0-(-10)=10\,\mathrm{K} \] Therefore, \[ Q=20\times2.1\times10 \] \[ Q=420\,\mathrm{J} \] \(\textbf{Identify the source of this heat:}\) Freezing liquid water releases latent heat. Let \(m_f\) be the mass of water that freezes. \[ Q=m_fL_f \] \[ 420=m_f\times334 \] \[ m_f=\frac{420}{334} \] \[ m_f=1.2575\,\mathrm{g} \] \(\textbf{Calculate the final ice mass:}\) The original added ice remains present, and an additional \(1.2575\,\mathrm{g}\) of liquid water freezes. \[ m_{\mathrm{ice,final}}=20+1.2575 \] \[ m_{\mathrm{ice,final}}=21.2575\,\mathrm{g} \] \[ m_{\mathrm{ice,final}}\approx21.26\,\mathrm{g} \] \(\textbf{Check the phase-equilibrium condition:}\) The remaining liquid water has mass \[ m_{\mathrm{water,final}}=100-1.2575 \] \[ m_{\mathrm{water,final}}=98.7425\,\mathrm{g} \] Both ice and liquid water therefore remain present. At atmospheric pressure, their equilibrium coexistence temperature is \(0^\circ\mathrm{C}\). The final total mass is \(120\,\mathrm{g}\), consistent with mass conservation. The essential inference is that adding cold ice to water at its freezing temperature can cause additional water to freeze rather than simply melting part of the added ice. \(\textbf{Final answer:}\) \[ \boxed{m_{\mathrm{ice,final}}\approx21.26\,\mathrm{g}} \]

Explanation: The response of solid-liquid equilibrium to pressure depends on the volume change accompanying melting. Increasing pressure favours the phase with the smaller molar volume. For substance X, the solid occupies less volume than the liquid. Therefore, increasing pressure favours its solid phase. At the original melting temperature, the higher pressure makes the solid relatively more stable. A higher temperature is required to restore solid-liquid coexistence. Consequently, the melting temperature of X increases. For substance Y, the solid occupies more volume than the liquid. Increasing pressure therefore favours the liquid phase. At the original melting temperature, melting becomes relatively more favourable. A lower temperature is required to restore coexistence. Thus, the melting temperature of Y decreases. The distinction explains why the pressure dependence of ice melting differs from that of many ordinary substances. For water near ordinary conditions, ice has a larger molar volume than liquid water and behaves like substance Y. The conclusion is based on the phase-volume relationship rather than on a universal rule that pressure always raises melting temperature. Hence, option C correctly predicts both responses.