201. A reversible reaction has \(K=10^{-3}\) at a certain temperature. The sign of \(\Delta G^\circ\) for the written reaction is
ⓐ. negative
ⓑ. positive
ⓒ. unrelated to \(K\)
ⓓ. always zero
Correct Answer: positive
Explanation: If \(K=10^{-3}\), then \(K\lt1\). For values of \(K\lt1\), \(\ln K\) is negative. In \(\Delta G^\circ=-RT\ln K\), the negative sign before \(RT\ln K\) changes the negative logarithm into a positive value. A positive \(\Delta G^\circ\) means the written reaction is reactant-favoured under standard-state comparison. This does not mean no product forms at equilibrium.
202. At \(298\,\text{K}\), a reaction has \(K=10\). Using \(R=8.314\,\text{J mol}^{-1}\text{K}^{-1}\) and \(\ln 10=2.303\), \(\Delta G^\circ\) is closest to
ⓐ. \(-24.8\,\text{kJ mol}^{-1}\)
ⓑ. \(+24.8\,\text{kJ mol}^{-1}\)
ⓒ. \(-5.71\,\text{kJ mol}^{-1}\)
ⓓ. \(+5.71\,\text{kJ mol}^{-1}\)
Correct Answer: \(-5.71\,\text{kJ mol}^{-1}\)
Explanation: \( \textbf{Relation used:} \)
\[\Delta G^\circ=-RT\ln K\]
\( \textbf{Given data:} \)
\(R=8.314\,\text{J mol}^{-1}\text{K}^{-1}\), \(T=298\,\text{K}\), \(K=10\), \(\ln 10=2.303\)
\( \textbf{Substitution:} \)
\[\Delta G^\circ=-(8.314)(298)(2.303)\,\text{J mol}^{-1}\]
\( \textbf{Intermediate multiplication:} \)
\[(8.314)(298)=2477.572\]
\[(2477.572)(2.303)\approx5706\,\text{J mol}^{-1}\]
\( \textbf{Sign and conversion:} \)
\[\Delta G^\circ\approx-5706\,\text{J mol}^{-1}=-5.71\,\text{kJ mol}^{-1}\]
\( \textbf{Final answer:} \) \(-5.71\,\text{kJ mol}^{-1}\). Since \(K\gt1\), the negative sign is consistent with a product-favoured written reaction.
203. Consider the following statements about \(K\) and \(\Delta G^\circ\):
I. A larger \(K\) generally corresponds to a more negative \(\Delta G^\circ\).
II. If \(K\lt1\), then \(\Delta G^\circ\) is positive for the written reaction.
III. At equilibrium, \(\Delta G=0\), but \(\Delta G^\circ\) need not be zero.
ⓐ. I and II only
ⓑ. I and III only
ⓒ. II and III only
ⓓ. I, II and III
Correct Answer: I, II and III
Explanation: Statement I follows from \(\Delta G^\circ=-RT\ln K\), because increasing \(K\) increases \(\ln K\) and makes \(\Delta G^\circ\) more negative. Statement II is true because \(K\lt1\) gives \(\ln K\lt0\), so \(\Delta G^\circ\gt0\). Statement III is also important because \(\Delta G\) and \(\Delta G^\circ\) are not the same quantity. At equilibrium, the actual Gibbs energy change \(\Delta G\) is zero. The standard value \(\Delta G^\circ\) depends on \(K\) and becomes zero only when \(K=1\).
204. A claim says, “At equilibrium, \(\Delta G^\circ\) must be zero because the reaction has no net change.” The better interpretation is that
ⓐ. \(\Delta G=0\) at equilibrium
ⓑ. \(K\) must always be equal to \(0\) at equilibrium
ⓒ. \(\Delta G\) and \(\Delta G^\circ\) are always identical
ⓓ. \(\Delta G^\circ=0\) for every reversible reaction
Correct Answer: \(\Delta G=0\) at equilibrium
Explanation: At equilibrium, the actual driving force for further net change is zero, so \(\Delta G=0\). The standard Gibbs energy change \(\Delta G^\circ\) is connected to the equilibrium constant by \(\Delta G^\circ=-RT\ln K\). Therefore \(\Delta G^\circ\) is zero only when \(K=1\). If \(K\gt1\), \(\Delta G^\circ\) is negative; if \(K\lt1\), it is positive. The word standard must not be ignored when interpreting the thermodynamic relation.
205. Le Chatelier principle is used to predict how an equilibrium system responds when
ⓐ. conditions are disturbed
ⓑ. all reacting particles stop moving permanently
ⓒ. the balanced equation is ignored
ⓓ. the equilibrium constant is removed from the reaction
Correct Answer: conditions are disturbed
Explanation: Le Chatelier principle describes the response of a system already at equilibrium when it is disturbed. The disturbance may involve concentration, pressure, volume, or temperature. The system shifts in a direction that tends to reduce the effect of the change. The principle predicts direction qualitatively, not the exact numerical composition. The equilibrium remains dynamic before and after the shift.
206. At fixed temperature, adding more reactant to a reversible reaction mixture usually causes the system to
ⓐ. make reactant and product concentrations equal
ⓑ. shift toward consuming the added reactant
ⓒ. change the value of \(K\) immediately
ⓓ. stop both forward and reverse reactions
Correct Answer: shift toward consuming the added reactant
Explanation: Adding a reactant increases its concentration immediately. The system responds by favouring the direction that uses up part of the added reactant. For a reactant written on the left side, this usually means a forward shift. At fixed temperature, the value of \(K\) does not change because \(K\) depends only on temperature for a given reaction. The new equilibrium composition is different, but it must still satisfy the same \(K\).
207. A product is removed from the equilibrium mixture \(\mathrm{A(g)+B(g)\rightleftharpoons C(g)}\) at constant temperature. The immediate response is best described as
ⓐ. no change because product concentration is irrelevant
ⓑ. a change in \(K_c\) without any shift
ⓒ. a shift toward reactants
ⓓ. a shift toward products
Correct Answer: a shift toward products
Explanation: Removing product lowers the product concentration term in the reaction quotient. For \(\mathrm{A(g)+B(g)\rightleftharpoons C(g)}\), this makes \(Q_c\) smaller than \(K_c\). The system then shifts forward to form more \(\mathrm{C}\). This partially replaces the removed product and helps restore the equilibrium relation. The equilibrium constant remains unchanged because the temperature has not changed.
208. At fixed temperature, a concentration change affects \(Q\) immediately, while \(K\)
ⓐ. changes only if the container is open
ⓑ. becomes zero until equilibrium returns
ⓒ. \(K\) remains constant at fixed temperature
ⓓ. changes in the same direction as \(Q\)
Correct Answer: \(K\) remains constant at fixed temperature
Explanation: The reaction quotient \(Q\) is calculated from the current concentrations or partial pressures. When a reactant or product is added or removed, the current composition changes immediately, so \(Q\) changes. The equilibrium constant \(K\) is fixed for a given reaction at a fixed temperature. The system shifts until \(Q\) again becomes equal to \(K\). Confusing \(Q\) with \(K\) can lead to a wrong prediction after concentration disturbance.
209. For \(\mathrm{A(g)+B(g)\rightleftharpoons C(g)}\), an equilibrium mixture has \(K_c=4.0\), \([\mathrm{A}]=0.50\,\text{M}\), \([\mathrm{B}]=0.50\,\text{M}\), and \([\mathrm{C}]=1.00\,\text{M}\). Extra \(\mathrm{A}\) is added suddenly so that \([\mathrm{A}]\) becomes \(1.00\,\text{M}\) before any shift occurs. The initial shift is
ⓐ. reverse
ⓑ. no net shift
ⓒ. forward
ⓓ. impossible to decide without changing \(K_c\)
Correct Answer: forward
Explanation: \( \textbf{Expression for the reaction:} \)
\[Q_c=\frac{[\mathrm{C}]}{[\mathrm{A}][\mathrm{B}]}\]
\( \textbf{Immediately after adding \(\mathrm{A}\):} \)
\[[\mathrm{A}]=1.00\,\text{M},\quad [\mathrm{B}]=0.50\,\text{M},\quad [\mathrm{C}]=1.00\,\text{M}\]
\( \textbf{Calculate the new quotient:} \)
\[Q_c=\frac{1.00}{(1.00)(0.50)}=2.0\]
\( \textbf{Compare with \(K_c\):} \)
\[Q_c=2.0,\quad K_c=4.0\]
\[Q_c\lt K_c\]
\( \textbf{Direction decision:} \)
When \(Q_c\lt K_c\), the system shifts forward to increase the product term.
\( \textbf{Final answer:} \) forward. Adding \(\mathrm{A}\) changes \(Q_c\) first; \(K_c\) remains \(4.0\) at the same temperature.
210. For \(\mathrm{A(g)\rightleftharpoons B(g)}\), a mixture is at equilibrium with \(K_c=3.0\). Some \(\mathrm{B}\) is removed suddenly. The system then shifts forward because
ⓐ. \(K_c\) becomes zero
ⓑ. \(Q_c\) becomes smaller than \(K_c\)
ⓒ. \(K_c\) becomes smaller than \(Q_c\)
ⓓ. \(Q_c\) becomes larger than \(K_c\)
Correct Answer: \(Q_c\) becomes smaller than \(K_c\)
Explanation: For \(\mathrm{A(g)\rightleftharpoons B(g)}\), \(Q_c=\frac{[\mathrm{B}]}{[\mathrm{A}]}\). Removing \(\mathrm{B}\) lowers the numerator of this expression. Therefore \(Q_c\) immediately becomes less than its equilibrium value \(K_c\). The system shifts forward to make more \(\mathrm{B}\) and raise \(Q_c\). The constant \(K_c\) is unchanged because the temperature is fixed.
211. A concentration-time graph for \(\mathrm{A\rightleftharpoons B}\) shows a sudden vertical increase in \([\mathrm{A}]\), followed by a gradual decrease in \([\mathrm{A}]\) and a gradual increase in \([\mathrm{B}]\). The disturbance most likely was
ⓐ. removal of \(\mathrm{A}\)
ⓑ. addition of a catalyst only
ⓒ. lowering the temperature with no concentration change
ⓓ. addition of \(\mathrm{A}\)
Correct Answer: addition of \(\mathrm{A}\)
Explanation: A sudden vertical jump in a concentration-time graph usually indicates direct addition of that species. Since \([\mathrm{A}]\) jumps upward, \(\mathrm{A}\) was added. After that, the system shifts forward, so \(\mathrm{A}\) is consumed gradually and \(\mathrm{B}\) is formed gradually. The gradual part represents the system moving toward a new equilibrium composition. A catalyst would not cause a sudden jump in the concentration of only one species.
212. Assertion: Adding a product to an equilibrium mixture at fixed temperature can make the reaction shift in the reverse direction.
Reason: Adding product makes the reaction quotient larger than the equilibrium constant for many product-over-reactant expressions.
ⓐ. Both assertion and reason are true, and the reason explains the assertion
ⓑ. The assertion is false, but the reason is true
ⓒ. The assertion is true, but the reason is false
ⓓ. Both assertion and reason are true, but the reason does not explain the assertion
Correct Answer: Both assertion and reason are true, and the reason explains the assertion
Explanation: The assertion is true because adding a product increases the product term in the reaction quotient. For many reactions written with products in the numerator, this makes \(Q\) greater than \(K\). When \(Q\gt K\), the system shifts in the reverse direction to reduce the product term. The reason therefore explains the direction of response. The value of \(K\) itself remains constant at fixed temperature.
213. A decrease in volume of a gaseous equilibrium mixture at constant temperature causes the system pressure to increase. The equilibrium tends to shift toward the side with
ⓐ. fewer number of gaseous moles
ⓑ. greater number of gaseous moles
ⓒ. equal masses of all species
ⓓ. greater number of solid moles
Correct Answer: fewer number of gaseous moles
Explanation: Decreasing volume increases pressure for a gas mixture. The system responds by favouring the side that has fewer moles of gas, because that direction helps reduce pressure. Only gaseous moles are counted for this pressure-shift prediction. Solids and liquids are not counted in the gaseous mole comparison. If the two sides have equal gaseous moles, a volume change does not cause a pressure-based shift.
214. For \(\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}\), increasing pressure by decreasing volume shifts the equilibrium
ⓐ. toward the side with more gas moles
ⓑ. toward \(\mathrm{N_2}\) and \(\mathrm{H_2}\)
ⓒ. neither way because all species are gases
ⓓ. toward \(\mathrm{NH_3}\)
Correct Answer: toward \(\mathrm{NH_3}\)
Explanation: The reactant side has \(1+3=4\) moles of gas. The product side has \(2\) moles of gas. Increasing pressure by compression favours the side with fewer gaseous moles. Therefore the equilibrium shifts toward \(\mathrm{NH_3}\). The fact that all species are gases does not remove the need to compare gaseous mole numbers.
215. For \(\mathrm{H_2(g)+I_2(g)\rightleftharpoons2HI(g)}\), compression at constant temperature produces no pressure-based shift because
ⓐ. the equilibrium constant becomes zero
ⓑ. gaseous mole numbers are equal
ⓒ. the reaction contains no gases
ⓓ. the product is a solid
Correct Answer: gaseous mole numbers are equal
Explanation: The reactant side has \(1+1=2\) moles of gas. The product side has \(2\) moles of gas. Since the gaseous mole numbers are equal, compression affects both sides in the same mole-count sense. There is no side that reduces pressure better than the other. The equilibrium constant also remains fixed at the same temperature.
216. The table gives pressure disturbances for gaseous equilibria.
| Case | Change | Reaction feature | Predicted shift |
|---|
| P | Compression | Products have fewer gaseous moles | Toward products |
| Q | Expansion | Products have more gaseous moles | Toward products |
| R | Compression | Both sides have equal gaseous moles | No pressure-based shift |
| S | Expansion | Reactants have fewer gaseous moles | Toward reactants |
The row that does not match the pressure-volume rule is
ⓐ. Q
ⓑ. P
ⓒ. S
ⓓ. R
Correct Answer: S
Explanation: Compression favours the side with fewer gaseous moles, so row \(P\) is suitable. Expansion favours the side with more gaseous moles, so row \(Q\) is suitable. If gaseous mole numbers are equal, there is no pressure-based shift, so row \(R\) is suitable. In row \(S\), expansion should favour the side with more gaseous moles, not the side with fewer gaseous moles. Since the reactants have fewer gaseous moles in row \(S\), the predicted shift should be away from reactants.
217. For \(\mathrm{PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)}\), increasing the volume at constant temperature shifts the equilibrium
ⓐ. left, because products are always disfavoured by expansion
ⓑ. neither way, because the reaction is reversible
ⓒ. right, because the right side has more gaseous moles
ⓓ. left, because the left side has fewer gaseous moles
Correct Answer: right, because the right side has more gaseous moles
Explanation: The left side has \(1\) mole of gaseous \(\mathrm{PCl_5}\). The right side has \(1+1=2\) moles of gaseous products. Increasing volume lowers pressure, so the system shifts toward the side with more gaseous moles. This means the equilibrium moves to the right. The prediction is based on gaseous mole count, not on the general word “product.”
218. Adding an inert gas to a gaseous equilibrium at constant volume usually causes no shift because
ⓐ. the total pressure remains unchanged
ⓑ. the inert gas reacts with all products
ⓒ. the equilibrium constant becomes larger
ⓓ. reacting partial pressures are unchanged
Correct Answer: reacting partial pressures are unchanged
Explanation: At constant volume and temperature, adding an inert gas increases the total pressure of the vessel. However, the amounts and volume of the reacting gases are unchanged, so their partial pressures remain the same. Since the reaction quotient for a gaseous equilibrium depends on reacting-gas partial pressures, \(Q_p\) does not change. The system therefore has no reason to shift. The total pressure alone is not enough; the partial pressures of reacting gases matter.
219. An inert gas is added to a gaseous equilibrium mixture at constant pressure. The volume increases, and the reacting gases are diluted. The shift, if any, is predicted by
ⓐ. treating it like an expansion and favouring the side with more gaseous moles
ⓑ. saying no shift is possible for any equilibrium
ⓒ. treating it like compression and favouring the side with fewer gaseous moles
ⓓ. changing \(K_p\) because an inert gas is present
Correct Answer: treating it like an expansion and favouring the side with more gaseous moles
Explanation: At constant pressure, adding an inert gas increases the volume of the system. The reacting gases become diluted, and their partial pressures decrease. This resembles expansion rather than compression. The equilibrium tends to shift toward the side with more gaseous moles if the two sides have different gaseous mole counts. The inert gas does not change \(K_p\) at fixed temperature.
220. For \(\mathrm{N_2O_4(g)\rightleftharpoons2NO_2(g)}\), lowering the pressure by increasing volume shifts the equilibrium
ⓐ. toward \(\mathrm{N_2O_4}\)
ⓑ. toward \(\mathrm{NO_2}\)
ⓒ. toward the side with fewer gas moles
ⓓ. neither way because both species contain nitrogen
Correct Answer: toward \(\mathrm{NO_2}\)
Explanation: The reactant side has \(1\) mole of gaseous \(\mathrm{N_2O_4}\). The product side has \(2\) moles of gaseous \(\mathrm{NO_2}\). Lowering pressure by increasing volume favours the side with more gaseous moles. Therefore the equilibrium shifts toward \(\mathrm{NO_2}\). The element composition is conserved, but the pressure response depends on gaseous mole numbers.
