101. The equilibrium \(\mathrm{Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)}\) has \(K_c\) equal to
ⓐ. \(\frac{[\mathrm{Fe^{3+}}]+[\mathrm{SCN^-}]}{[\mathrm{FeSCN^{2+}}]}\)
ⓑ. \([\mathrm{FeSCN^{2+}}][\mathrm{Fe^{3+}}][\mathrm{SCN^-}]\)
ⓒ. \(\frac{[\mathrm{FeSCN^{2+}}]}{[\mathrm{Fe^{3+}}][\mathrm{SCN^-}]}\)
ⓓ. \(\frac{[\mathrm{Fe^{3+}}][\mathrm{SCN^-}]}{[\mathrm{FeSCN^{2+}}]}\)
Correct Answer: \(\frac{[\mathrm{FeSCN^{2+}}]}{[\mathrm{Fe^{3+}}][\mathrm{SCN^-}]}\)
Explanation: All species in this equilibrium are aqueous, so each one can appear in the \(K_c\) expression. The product \(\mathrm{FeSCN^{2+}}\) is placed in the numerator. The reactants \(\mathrm{Fe^{3+}}\) and \(\mathrm{SCN^-}\) are placed in the denominator. Since all coefficients are \(1\), no concentration term needs an exponent greater than \(1\). Charge notation must remain attached to the correct species while writing the expression.
102. A \(K_c\) expression for \(\mathrm{2A(aq)+B(aq)\rightleftharpoons3C(aq)}\) is being checked. The exponent on \([\mathrm{C}]\) must be \(3\) because
ⓐ. aqueous species always receive exponent \(3\)
ⓑ. \(\mathrm{C}\) is a product with coefficient \(3\)
ⓒ. \(\mathrm{C}\) is the third letter in the equation
ⓓ. there are three different species in the reaction
Correct Answer: \(\mathrm{C}\) is a product with coefficient \(3\)
Explanation: The power of a concentration term in \(K_c\) comes from the coefficient of that species in the balanced equation. Since \(\mathrm{C}\) has coefficient \(3\), the product term is \([\mathrm{C}]^3\). The exponent is not related to the letter name or the total number of species present. The aqueous phase label tells that the species may be included, but it does not decide the exponent. Coefficients and phases play different roles while writing \(K_c\).
103. For a fixed balanced reaction, the value of \(K_c\) changes when
ⓐ. the temperature is changed
ⓑ. the same equilibrium mixture is placed in a darker room
ⓒ. the initial concentration is changed at the same temperature
ⓓ. a catalyst is added at the same temperature
Correct Answer: the temperature is changed
Explanation: The equilibrium constant \(K_c\) is constant for a given balanced reaction at a fixed temperature. Changing initial concentration can change the path to equilibrium and the final composition, but not the value of \(K_c\) at that temperature. A catalyst helps the system reach equilibrium faster but does not change \(K_c\). Temperature changes can change the relative tendency of forward and reverse processes. That is why the temperature condition must be stated when giving an equilibrium constant.
104. A data record gives two equilibrium mixtures for the same reaction \(\mathrm{A(g)\rightleftharpoons B(g)}\) at the same temperature.
| Mixture | \([\mathrm{A}]\) | \([\mathrm{B}]\) |
|---|
| P | \(0.20\,\text{M}\) | \(0.60\,\text{M}\) |
| Q | \(0.40\,\text{M}\) | \(1.20\,\text{M}\) |
The best conclusion is that
ⓐ. only mixture Q can be at equilibrium
ⓑ. neither can be at equilibrium because \([\mathrm{A}]\neq[\mathrm{B}]\)
ⓒ. only mixture P can be at equilibrium
ⓓ. both give the same \(K_c\) ratio
Correct Answer: both give the same \(K_c\) ratio
Explanation: \( \textbf{Reaction:} \)
\[
\mathrm{A(g)\rightleftharpoons B(g)}
\]
\( \textbf{Equilibrium expression:} \)
\[
K_c=\frac{[\mathrm{B}]}{[\mathrm{A}]}
\]
\( \textbf{For mixture P:} \)
\[
K_c=\frac{0.60}{0.20}=3.0
\]
\( \textbf{For mixture Q:} \)
\[
K_c=\frac{1.20}{0.40}=3.0
\]
\( \textbf{Comparison:} \)
Both mixtures give the same value of \(K_c\).
\( \textbf{Interpretation:} \)
Different equilibrium concentrations can still satisfy the same equilibrium constant at the same temperature.
\( \textbf{Final answer:} \) both can have the same \(K_c\). Equal concentrations are not required; the required relation is the one set by the equilibrium expression.
105. A value of \(K_c\) much greater than \(1\) for a reversible reaction usually indicates that, at equilibrium,
ⓐ. products are favoured over reactants
ⓑ. reactants and products must have equal concentrations
ⓒ. the forward reaction has stopped completely
ⓓ. the reaction is impossible in the reverse direction
Correct Answer: products are favoured over reactants
Explanation: The value of \(K_c\) compares the concentration terms of products with those of reactants in the equilibrium expression. When \(K_c\) is much greater than \(1\), the numerator is relatively large compared with the denominator. This means the equilibrium mixture is product-favoured for the written reaction. It does not mean that all reactants have disappeared. A large \(K_c\) still describes a dynamic equilibrium in which both directions can continue.
106. For a reaction with \(K_c\ll1\), the safest conclusion about the equilibrium mixture is that
ⓐ. it contains no products at all
ⓑ. it is reactant-favoured
ⓒ. it is not a reversible reaction
ⓓ. it must have equal forward and reverse concentrations
Correct Answer: it is reactant-favoured
Explanation: A very small \(K_c\) means the product concentration terms are small compared with the reactant concentration terms at equilibrium. The written reaction therefore favours reactants under those conditions. This does not mean that products are completely absent. Even a small amount of product can be present at equilibrium. The statement is about relative extent, not about the reaction stopping.
107. For \(\mathrm{A(g)\rightleftharpoons B(g)}\), \(K_c=4.0\) at a certain temperature. If \([\mathrm{A}]=0.25\,\text{M}\) at equilibrium, the equilibrium concentration of \(\mathrm{B}\) is
ⓐ. \(4.25\,\text{M}\)
ⓑ. \(0.25\,\text{M}\)
ⓒ. \(1.0\,\text{M}\)
ⓓ. \(0.0625\,\text{M}\)
Correct Answer: \(1.0\,\text{M}\)
Explanation: \( \textbf{Reaction considered:} \)
\[\mathrm{A(g)\rightleftharpoons B(g)}\]
\( \textbf{Equilibrium expression:} \)
\[K_c=\frac{[\mathrm{B}]}{[\mathrm{A}]}\]
\( \textbf{Given values:} \)
\(K_c=4.0\) and \([\mathrm{A}]=0.25\,\text{M}\)
\( \textbf{Substitution into the relation:} \)
\[4.0=\frac{[\mathrm{B}]}{0.25}\]
\( \textbf{Solving for product concentration:} \)
\[[\mathrm{B}]=4.0\times0.25\]
\[[\mathrm{B}]=1.0\,\text{M}\]
\( \textbf{Reporting check:} \)
The result is a concentration, so the unit is \(\text{M}\) or \(\text{mol L}^{-1}\).
\( \textbf{Final answer:} \) \(1.0\,\text{M}\). The value \(K_c=4.0\) means \([\mathrm{B}]\) is four times \([\mathrm{A}]\) for this one-to-one equilibrium.
108. The unit of \(K_c\) for \(\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}\), when concentration is reported in \(\text{mol L}^{-1}\), is
ⓐ. \(\text{mol}^2\text{L}^{-2}\)
ⓑ. no unit because every \(K_c\) is unitless
ⓒ. \(\text{mol L}^{-1}\)
ⓓ. \(\text{L}^2\text{mol}^{-2}\)
Correct Answer: \(\text{L}^2\text{mol}^{-2}\)
Explanation: \( \textbf{Balanced reaction:} \)
\[\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}\]
\( \textbf{Equilibrium expression:} \)
\[K_c=\frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}\]
\( \textbf{Concentration unit used:} \)
Each concentration has unit \(\text{mol L}^{-1}\).
\( \textbf{Unit in numerator:} \)
\[(\text{mol L}^{-1})^2\]
\( \textbf{Unit in denominator:} \)
\[(\text{mol L}^{-1})(\text{mol L}^{-1})^3=(\text{mol L}^{-1})^4\]
\( \textbf{Net unit:} \)
\[\frac{(\text{mol L}^{-1})^2}{(\text{mol L}^{-1})^4}=(\text{mol L}^{-1})^{-2}\]
\[(\text{mol L}^{-1})^{-2}=\text{L}^2\text{mol}^{-2}\]
\( \textbf{Final answer:} \) \(\text{L}^2\text{mol}^{-2}\). The unit depends on the total powers in the written equilibrium expression.
109. The unit of \(K_c\) becomes absent in the concentration-unit method for a reaction such as
ⓐ. \(\mathrm{2SO_2(g)+O_2(g)\rightleftharpoons2SO_3(g)}\)
ⓑ. \(\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}\)
ⓒ. \(\mathrm{H_2(g)+I_2(g)\rightleftharpoons2HI(g)}\)
ⓓ. \(\mathrm{PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)}\)
Correct Answer: \(\mathrm{H_2(g)+I_2(g)\rightleftharpoons2HI(g)}\)
Explanation: For \(\mathrm{H_2(g)+I_2(g)\rightleftharpoons2HI(g)}\), the expression is \(K_c=\frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}\). The total concentration power in the numerator is \(2\), and the total concentration power in the denominator is also \(2\). Therefore the concentration units cancel in the unit calculation. In the other listed reactions, the total powers on the two sides of the expression are not equal. Equal total powers in the expression are the reason the unit cancels here.
110. A table gives possible meanings of \(K_c\) for the same reaction at fixed temperature.
| Case | Claim about \(K_c\) |
|---|
| P | It is calculated from equilibrium concentrations. |
| Q | It changes whenever a catalyst is added. |
| R | Its magnitude helps judge whether products or reactants are favoured. |
| S | It is written without using the balanced equation. |
The two acceptable claims are
ⓐ. R and S
ⓑ. P and R
ⓒ. P and Q
ⓓ. Q and S
Correct Answer: P and R
Explanation: \(K_c\) is obtained from the equilibrium concentrations placed in the expression for the balanced reaction. Its magnitude gives useful information about whether the written reaction is product-favoured or reactant-favoured. A catalyst does not change \(K_c\) at a fixed temperature; it only helps the system reach equilibrium faster. The balanced equation is necessary because coefficients become powers in the expression. Claims \(P\) and \(R\) describe the role of \(K_c\) most reliably.
111. A reversible reaction has \(K_c=1.0\) at a fixed temperature. This value most directly suggests that
ⓐ. the reaction has no forward or reverse movement
ⓑ. all individual concentrations must be exactly \(1.0\,\text{M}\)
ⓒ. the reaction can never respond to disturbance
ⓓ. product and reactant terms are comparable
Correct Answer: product and reactant terms are comparable
Explanation: A \(K_c\) value near \(1\) means the numerator and denominator of the equilibrium expression are comparable. This does not require every concentration in the mixture to be \(1.0\,\text{M}\). It also does not demand that each reactant concentration equals each product concentration. The actual species concentrations depend on the balanced equation and the powers in the expression. The value \(K_c=1.0\) describes a ratio of concentration terms, not a universal concentration value.
112. For \(\mathrm{CH_3COOH(aq)+C_2H_5OH(aq)\rightleftharpoons CH_3COOC_2H_5(aq)+H_2O(l)}\), a \(K_c\) expression that omits the pure liquid water term is
ⓐ. \(K_c=\frac{[\mathrm{CH_3COOC_2H_5}]}{[\mathrm{CH_3COOH}][\mathrm{C_2H_5OH}]}\)
ⓑ. \(K_c=[\mathrm{CH_3COOH}]+[\mathrm{C_2H_5OH}]+[\mathrm{CH_3COOC_2H_5}]\)
ⓒ. \(K_c=\frac{[\mathrm{CH_3COOH}][\mathrm{C_2H_5OH}]}{[\mathrm{CH_3COOC_2H_5}]}\)
ⓓ. \(K_c=\frac{[\mathrm{CH_3COOC_2H_5}][\mathrm{H_2O}]}{[\mathrm{CH_3COOH}][\mathrm{C_2H_5OH}]}\)
Correct Answer: \(K_c=\frac{[\mathrm{CH_3COOC_2H_5}]}{[\mathrm{CH_3COOH}][\mathrm{C_2H_5OH}]}\)
Explanation: In this written equilibrium, ester \(\mathrm{CH_3COOC_2H_5}\) is a product and appears in the numerator. Acetic acid \(\mathrm{CH_3COOH}\) and ethanol \(\mathrm{C_2H_5OH}\) are reactants and appear in the denominator. The pure liquid water term is omitted in the expression because its active mass is treated as constant. All shown coefficients are \(1\), so no power greater than \(1\) is needed. The expression must use multiplication of concentration terms, not addition.
113. A weak electrolyte ionizes as \(\mathrm{HA(aq)\rightleftharpoons H^+(aq)+A^-(aq)}\). Its equilibrium expression is
ⓐ. \(K_c=\frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}\)
ⓑ. \(K_c=\frac{[\mathrm{H^+}]}{[\mathrm{A^-}]}\)
ⓒ. \(K_c=\frac{[\mathrm{HA}]}{[\mathrm{H^+}][\mathrm{A^-}]}\)
ⓓ. \(K_c=[\mathrm{H^+}]+[\mathrm{A^-}]-[\mathrm{HA}]\)
Correct Answer: \(K_c=\frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}\)
Explanation: The products of ionization are \(\mathrm{H^+}\) and \(\mathrm{A^-}\), so their concentration terms are placed in the numerator. The undissociated weak electrolyte \(\mathrm{HA}\) is the reactant and appears in the denominator. All coefficients are \(1\), so no exponent greater than \(1\) appears. This expression is also commonly written as \(K_a\) when \(\mathrm{HA}\) is treated as a weak acid. The charges must stay attached to the ions because \(\mathrm{H^+}\), \(\mathrm{A^-}\), and \(\mathrm{HA}\) are different species.
114. For \(\mathrm{C(s)+CO_2(g)\rightleftharpoons2CO(g)}\), the proper \(K_c\) expression is
ⓐ. \(K_c=\frac{[\mathrm{C}][\mathrm{CO_2}]}{[\mathrm{CO}]^2}\)
ⓑ. \(K_c=\frac{[\mathrm{CO}]^2}{[\mathrm{CO_2}]}\)
ⓒ. \(K_c=\frac{[\mathrm{CO}]^2}{[\mathrm{C}][\mathrm{CO_2}]}\)
ⓓ. \(K_c=\frac{2[\mathrm{CO}]}{[\mathrm{CO_2}]}\)
Correct Answer: \(K_c=\frac{[\mathrm{CO}]^2}{[\mathrm{CO_2}]}\)
Explanation: The product \(\mathrm{CO(g)}\) has coefficient \(2\), so its concentration term is \([\mathrm{CO}]^2\). The gaseous reactant \(\mathrm{CO_2(g)}\) appears in the denominator. Solid carbon \(\mathrm{C(s)}\) is omitted because it is a pure solid with constant active mass. The coefficient \(2\) becomes an exponent, not a multiplier. The final expression keeps only variable gaseous concentration terms from the balanced equation.
115. A proposed expression for \(\mathrm{MgCO_3(s)\rightleftharpoons MgO(s)+CO_2(g)}\) is \(K_c=\frac{[\mathrm{MgO}][\mathrm{CO_2}]}{[\mathrm{MgCO_3}]}\). The best repair is
ⓐ. \(K_c=\frac{[\mathrm{MgCO_3}]}{[\mathrm{MgO}][\mathrm{CO_2}]}\)
ⓑ. \(K_c=\frac{1}{[\mathrm{CO_2}]}\)
ⓒ. \(K_c=[\mathrm{MgO}][\mathrm{CO_2}]\)
ⓓ. \(K_c=[\mathrm{CO_2}]\)
Correct Answer: \(K_c=[\mathrm{CO_2}]\)
Explanation: Both \(\mathrm{MgCO_3(s)}\) and \(\mathrm{MgO(s)}\) are pure solids. Pure solids are not written in the \(K_c\) expression because their active masses remain constant. The only variable species shown in the equation is \(\mathrm{CO_2(g)}\). Therefore the expression reduces to \(K_c=[\mathrm{CO_2}]\). The solids still participate in the equilibrium, but their concentration terms are not included.
116. A reaction is written as \(\mathrm{A(g)+B(g)\rightleftharpoons3C(g)}\). If concentration is measured in \(\text{mol L}^{-1}\), the unit of \(K_c\) for this reaction is
ⓐ. \(\text{L mol}^{-1}\)
ⓑ. no unit by cancellation
ⓒ. \(\text{mol L}^{-1}\)
ⓓ. \(\text{L}^2\text{mol}^{-2}\)
Correct Answer: \(\text{mol L}^{-1}\)
Explanation: \( \textbf{Balanced reaction:} \)
\[
\mathrm{A(g)+B(g)\rightleftharpoons3C(g)}
\]
\( \textbf{Equilibrium expression:} \)
\[
K_c=\frac{[\mathrm{C}]^3}{[\mathrm{A}][\mathrm{B}]}
\]
\( \textbf{Unit of each concentration term:} \)
\[
[\text{concentration}]=\text{mol L}^{-1}
\]
\( \textbf{Numerator unit:} \)
\[
(\text{mol L}^{-1})^3
\]
\( \textbf{Denominator unit:} \)
\[
(\text{mol L}^{-1})(\text{mol L}^{-1})=(\text{mol L}^{-1})^2
\]
\( \textbf{Net unit:} \)
\[
\frac{(\text{mol L}^{-1})^3}{(\text{mol L}^{-1})^2}=\text{mol L}^{-1}
\]
\( \textbf{Final answer:} \) \(\text{mol L}^{-1}\). The unit does not cancel because the total concentration power in the numerator is greater than that in the denominator by \(1\).
117. For \(\mathrm{2A(g)+B(g)\rightleftharpoons C(g)+3D(g)}\), with all concentrations in \(\text{mol L}^{-1}\), the unit of \(K_c\) is
ⓐ. \(\text{mol L}^{-1}\)
ⓑ. \(\text{L}^2\text{mol}^{-2}\)
ⓒ. no unit concentration-unit method
ⓓ. \(\text{L mol}^{-1}\)
Correct Answer: \(\text{mol L}^{-1}\)
Explanation: \( \textbf{Equilibrium expression:} \)
\[K_c=\frac{[\mathrm{C}][\mathrm{D}]^3}{[\mathrm{A}]^2[\mathrm{B}]}\]
\( \textbf{Unit of each concentration:} \)
\[\text{mol L}^{-1}\]
\( \textbf{Numerator unit:} \)
\[(\text{mol L}^{-1})(\text{mol L}^{-1})^3=(\text{mol L}^{-1})^4\]
\( \textbf{Denominator unit:} \)
\[(\text{mol L}^{-1})^2(\text{mol L}^{-1})=(\text{mol L}^{-1})^3\]
\( \textbf{Net unit:} \)
\[\frac{(\text{mol L}^{-1})^4}{(\text{mol L}^{-1})^3}=\text{mol L}^{-1}\]
\( \textbf{Final answer:} \) \(\text{mol L}^{-1}\). The unit follows from the difference between total powers in numerator and denominator.
118. The pressure equilibrium constant \(K_p\) is most suitable when the equilibrium expression is written using
ⓐ. masses of pure solids only
ⓑ. colour intensities of products
ⓒ. gaseous partial pressures
ⓓ. molar masses of all species
Correct Answer: gaseous partial pressures
Explanation: \(K_p\) is the equilibrium constant expressed in terms of partial pressures. It is used for gaseous equilibria because each gas in a mixture exerts its own partial pressure. The powers in a \(K_p\) expression still come from the balanced equation. Pure solids and pure liquids are not represented by pressure terms. The symbol \(K_p\) therefore points to gaseous species and their partial pressures, not to mass or colour.
119. For \(\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}\), the expression for \(K_p\) is
ⓐ. \(K_p=\frac{p_{\mathrm{N_2}}(p_{\mathrm{H_2}})^3}{(p_{\mathrm{NH_3}})^2}\)
ⓑ. \(K_p=p_{\mathrm{N_2}}+p_{\mathrm{H_2}}+p_{\mathrm{NH_3}}\)
ⓒ. \(K_p=\frac{2p_{\mathrm{NH_3}}}{p_{\mathrm{N_2}}+3p_{\mathrm{H_2}}}\)
ⓓ. \(K_p=\frac{(p_{\mathrm{NH_3}})^2}{p_{\mathrm{N_2}}(p_{\mathrm{H_2}})^3}\)
Correct Answer: \(K_p=\frac{(p_{\mathrm{NH_3}})^2}{p_{\mathrm{N_2}}(p_{\mathrm{H_2}})^3}\)
Explanation: The \(K_p\) expression uses partial pressures of gaseous species. \(\mathrm{NH_3}\) is the product and has coefficient \(2\), so \((p_{\mathrm{NH_3}})^2\) appears in the numerator. \(\mathrm{N_2}\) and \(\mathrm{H_2}\) are reactants, so their pressure terms appear in the denominator. The coefficient \(3\) of \(\mathrm{H_2}\) becomes the power on \(p_{\mathrm{H_2}}\). Partial pressure terms are multiplied according to the balanced equation, not added.
120. In \(K_p\) expressions, a dissolved species such as \(\mathrm{Cl^-(aq)}\) is not written as a pressure term because
ⓐ. aqueous species must always be placed in the numerator
ⓑ. \(K_p\) uses partial pressures only for gaseous species
ⓒ. ions cannot take part in equilibrium
ⓓ. all dissolved ions have zero concentration
Correct Answer: \(K_p\) uses partial pressures only for gaseous species
Explanation: Partial pressure is a gas-mixture quantity. A dissolved ion such as \(\mathrm{Cl^-(aq)}\) belongs to the aqueous phase, so it is not represented by \(p_{\mathrm{Cl^-}}\). If an equilibrium contains aqueous species, their concentrations may appear in a \(K_c\)-type expression instead. \(K_p\) is restricted to gaseous species in the pressure expression. The absence of an aqueous pressure term does not mean the ion is chemically unimportant.
