Equilibrium – NEET UG ChemistryPractice Questions & PYQs

To determine the equilibrium constant for a reversible reaction, one must calculate the final concentrations of all species by applying the principles of stoichiometry to the reaction progress. Given that the reaction starts with 1.00 M of both reactants and reaches equilibrium with 0.25 M of product D, the stoichiometric coefficients indicate the extent of the reaction. Because the coefficient of D is 1, the formation of 0.25 M of D implies that the concentration of C, which has a coefficient of 3, must increase by 0.75 M, while reactants A and B, with coefficients of 2 and 1, are consumed by 0.50 M and 0.25 M, respectively.

Calculating the equilibrium concentrations involves subtracting these consumption values from the initial reactant amounts and adding the formation values to the products. This yields final concentrations of 0.50 M for A, 0.75 M for B, 0.75 M for C, and 0.25 M for D. Substituting these values into the equilibrium constant expression, which is defined as the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients, results in the following equation:

$K = \frac{[\mathrm{C}]^3[\mathrm{D}]}{[\mathrm{A}]^2[\mathrm{B}]} = \frac{(0.75)^3(0.25)}{(0.50)^2(0.75)}$

For a basic buffer composed of a weak base and its conjugate acid, the relationship between the hydroxide ion concentration and the buffer components is governed by the Henderson-Hasselbalch equation, $pOH = pK_b + \log \frac{[HB]}{[B^-]}$}$. When the concentrations of the base$B^-$and its conjugate acid$HB$are equal, the ratio \in the logarithmic term becomes one, causing the entire logarithmic term to vanish since$\log(1)$equals zero. This simplifies the relationship to$pOH = pK_b$. Given the dissociation constant$K_b$of$10^{-10}$, we find$pK_b$ by taking the negative logarithm of this value, yielding 10. Consequently, since the sum of pH and pOH in an aqueous solution at room temperature is 14, subtracting the calculated pOH of 10 from 14 results in a final pH of 4.

The relationship between equilibrium constants $K_p$ and $K_c$ is defined by the equation $K_p = K_c(RT)^{\Delta n}$, where $\Delta n$ represents the difference between the total stoichiometric coefficients of gaseous products and gaseous reactants. When $\Delta n$ is equal to zero, the term $(RT)^0$ simplifies to unity, making the two constants identical. Conversely, if $\Delta n$ is not zero, the term $(RT)^{\Delta n}$ ensures that $K_p$ and $K_c$ differ in value.

In the reaction $2\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{CO}_2(\mathrm{g})$, we must focus exclusively on the gaseous species and ignore the solid carbon. This leaves us with two moles of gaseous $\mathrm{CO}_2$ as products and one mole of gaseous $\mathrm{O}_2$ as a reactant. Determining the difference yields $\Delta n = 2 - 1 = 1$, resulting in the expression $K_p = K_c(RT)^1$. Because the exponent is non-zero, the values of $K_p$ and $K_c$ are not equal.

A buffer solution consisting of a weak acid, acetic acid, and its conjugate base, provided by sodium acetate, maintains equilibrium concentrations approximately equal to the initial concentrations due to the common ion effect. Since acetic acid is a weak acid, its dissociation is negligible, meaning the concentration of undissociated acid remains at 0.10 M and the concentration of acetate ions remains at 0.20 M. The hydrogen ion concentration is determined by the equilibrium constant expression:

[\mathrm{H}^{+}] = K_a \frac{[\mathrm{CH}_3\mathrm{COOH}]}{[\mathrm{CH}_3\mathrm{COO}^{-}]}$}

Substituting the given values into this expression results in $[\mathrm{H}^{+}] = 1.8 \times 10^{-5} \times \frac{0.10}{0.20}$, which simplifies to $9.0 \times 10^{-6}$ M.

The solubility product constant describes the equilibrium of a saturated solution based on the concentrations of its dissociated ions. Given a pH of 12, the pOH is 2, which corresponds to a hydroxide ion concentration of $10^{-2}$ M. Dissociation of $\mathrm{Ba}(\mathrm{OH})_{2}$ produces one barium ion for every two hydroxide ions, meaning the hydroxide concentration equals $2s$ where $s$ represents the molar solubility. Solving $2s = 10^{-2}$ gives a solubility of $s = 0.5 \times 10^{-2}$ M.

The equilibrium expression for this process is represented by the relationship:
K_{sp} = $[\mathrm{Ba}^{2+}]\[\mathrm{OH}^{-}]^2 = 4s^3$$
Substituting the calculated solubility into this formula produces $4 \times (0.5 \times 10^{-2})^3$, which results in a value of $5.00 \times 10^{-7}$.

The reaction between $\text{HCl}$ and $\text{Ba}(\text{OH})_2$ is a neutralization process where $\text{H}^+$ and $\text{OH}^-$ ions combine to form water, leaving behind an excess of whichever species was initially present in a greater quantity. $\text{HCl}$ acts as a monoprotic acid that dissociates completely to provide one $\text{H}^+$ ion per molecule, while $\text{Ba}(\text{OH})_2$ is a strong base that dissociates to release two $\text{OH}^-$ ions for every molecule.

The amount of acid is calculated as $20 \times 0.05 = 1$ mmol of $\text{H}^+$. The total hydroxide contribution is determined by multiplying the base volume, molarity, and the factor of two, resulting in $30 \times 0.1 \times 2 = 6$ mmol of $\text{OH}^-$. Because the base is in excess, the difference of $6 - 1 = 5$ mmol represents the remaining hydroxide ions. Dividing this excess by the total volume of $20 + 30 = 50$ mL yields a final hydroxide ion concentration of $5 / 50 = 0.1$ M.

A Lewis acid is defined as a chemical species capable of accepting an electron pair, a behavior typical of electron-deficient molecules that have not achieved a complete octet. In the case of $(CH_3)_3B$, the central Boron atom forms three covalent bonds but remains electron-deficient with only six electrons in its valence shell. Because of this incomplete octet, the molecule acts as a Lewis acid, whereas the other options contain lone pairs on the central atom and therefore function as Lewis bases.

Ammonium chloride acts as a salt composed of a strong acid and a weak base, meaning we determine its hydrolysis constant by relating the dissociation constant of the weak base to the ionic product of water. The relationship is defined by the equation $K_h = \frac{K_w}{K_b}$ where $K_w = 10^{-14}$ is the ionic product of water at 298 K and $K_b = 1.77 × 10^{-5}$ is the ionization constant for ammonium hydroxide. By substituting these values into the expression, we perform the calculation $K_h = \frac{10^{-14}}{1.77 \times 10^{-5}}$ which simplifies to approximately $0.565 × 10^{-9}$. Adjusting the scientific notation yields a final value of $5.65 × 10^{-10}$.

The overall equilibrium constant for a chemical reaction is determined by manipulating the equilibrium expressions of the individual steps that compose the overall process. In this case, the reaction $\mathrm{CN}^{-}+\mathrm{CH}_3\mathrm{COOH} \rightleftharpoons \mathrm{HCN}+\mathrm{CH}_3\mathrm{COO}^{-}$ can be represented as the sum of the dissociation of acetic acid, given by $K_1 = 1.5 \times 10^{-5}$, and the reverse dissociation of hydrocyanic acid, which has an equilibrium constant equal to the reciprocal of its dissociation constant, $1/K_2$ with $K_2 = 4.5 \times 10^{-10}$.

Mathematically, the total equilibrium constant $K$ for this transformation is expressed as the ratio of the two dissociation constants, $K = \frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}}$. Simplifying the powers of ten yields $10^5$, and dividing the coefficients $1.5$ by $4.5$ results in approximately $0.3$, producing a final value of $3.0 \times 10^4$.

When manipulating chemical equations, the equilibrium constant adjusts mathematically based on the changes made to the reaction stoichiometry. Reversing a balanced chemical equation results in a new equilibrium constant that is the reciprocal of the original value. Additionally, multiplying the coefficients of an entire equation by a specific factor requires raising the equilibrium constant to the power of that factor. Reversing the reaction $\mathrm{HI}(\mathrm{g})\rightleftharpoons\frac{1}{2}\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2}\mathrm{I}_{2}(\mathrm{g})$ changes the equilibrium constant from 8 to 1/8. Because the target reaction $\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g})\rightleftharpoons 2\mathrm{HI}(\mathrm{g})$ effectively doubles the coefficients of the reversed equation, you calculate the new equilibrium constant by squaring 1/8, which results in 1/64.

Applying the law of mass action to the dissociation equilibrium
$2\mathrm{AB}_2(g)\rightleftharpoons 2\mathrm{AB}(g)+\mathrm{B}_2(g)$
allows us to determine the relationship between the equilibrium constant $K_P$, total pressure $P$, and degree of dissociation $x$. Starting with 2 moles of $\mathrm{AB}_2$, the amounts at equilibrium are $2(1-x)$, $2x$, and $x$ for $\mathrm{AB}_2$, $\mathrm{AB}$, and $\mathrm{B}_2$ respectively, which leads to a total of $2+x$ moles. Because $x$ is small, we can approximate $2+x$ as 2 and the term $(1-x)$ as 1, which results in the partial pressures $P_{\mathrm{AB}_2} \approx P$, $P_{\mathrm{AB}} \approx xP$, and $P_{\mathrm{B}_2} \approx \frac{x}{2}P$. Substituting these into the equilibrium expression $K_P = \frac{P_{\mathrm{AB}}^2 P_{\mathrm{B}_2}}{P_{\mathrm{AB}_2}^2}$ gives:
$K_P = \frac{(xP)^2 (\frac{x}{2}P)}{P^2}$
Simplifying this equation leads to $K_P = \frac{x^3P}{2}$, which can be rearranged to provide the final relationship:
$x = \left(\frac{2K_p}{P}\right)^{1/3}$

The equilibrium constant $K_p$ for a gaseous reaction is defined by the partial pressures of its constituents, which are determined by the total system pressure and the extent of chemical dissociation. For the decomposition reaction $X \rightleftharpoons Y + Z$, starting with 1 mole and a degree of dissociation $\alpha$, the equilibrium mixture contains $1-\alpha$ moles of $X$ and $\alpha$ moles each of $Y$ and $Z$, totaling $1+\alpha$ moles. Expressing partial pressures as the mole fraction multiplied by the total pressure $P_1$, the equilibrium constant is defined as
$K_{p_1} = \frac{\alpha^2 P_1}{1-\alpha^2}$

For the second reaction $A \rightleftharpoons 2B$, starting with 1 mole and the same degree of dissociation $\alpha$, the equilibrium mixture consists of $1-\alpha$ moles of $A$ and $2\alpha$ moles of $B$, also totaling $1+\alpha$ moles. This gives the equilibrium constant expression
$K_{p_2} = \frac{4\alpha^2 P_2}{1-\alpha^2}$
Setting the ratio of these constants to $K_{p_1} / K_{p_2} = 9$ cancels the terms containing $\alpha$ entirely, which simplifies the relationship to $\frac{P_1}{4P_2} = 9$. Consequently, the ratio of total pressures $P_1$ to $P_2$ is 36:1.

Concentration in a mixture depends on dividing the total moles of solute by the combined volume of the solution. Assign a variable $V$ to represent the equal volume of each acid component. Use the relationship $[H^{+}] = 10^{-pH}$ to convert the provided pH values into hydrogen ion concentrations of $10^{-3}$ M, $10^{-4}$ M, and $10^{-5}$ M. The total amount of $H^{+}$ ions is found by summing the individual contributions, yielding $V(10^{-3} + 10^{-4} + 10^{-5})$, while the final volume is $3V$. Dividing the total moles by the total volume provides the final concentration:

$\frac{V(10^{-3} + 10^{-4} + 10^{-5})}{3V} = \frac{10^{-3} + 10^{-4} + 10^{-5}}{3}$

Expressing the sum $10^{-3} + 10^{-4} + 10^{-5}$ as $10^{-3}(1 + 0.1 + 0.01)$ results in $1.11 \times 10^{-3}$. Dividing this figure by $3$ simplifies to $0.37 \times 10^{-3}$, which is equivalent to $3.7 \times 10^{-4}$ M.

The equilibrium constant for a chemical reaction remains invariant at a given temperature, which necessitates that the product of the concentrations of the species involved, raised to their respective stoichiometric coefficients, must remain constant. For the reaction $\mathrm{Fe}(\mathrm{OH})_{3} \rightleftharpoons \mathrm{Fe}^{3+} + 3\mathrm{OH}^{-}$, the equilibrium expression is given by $K =$[\mathrm{Fe}^{3+}]$$[\mathrm{OH}^{-}]^3$, with the solid phase excluded from the equation because its activity is treated as unity. When the concentration of hydroxide ions is reduced to $1/4$ of its original value, the cubic term \in the expression becomes $(1/4)^3$, which equals $1/64$. To maintain the constant value of $K$, the concentration of $\mathrm{Fe}^{3+}$ ions must consequently increase by a factor of 64.

For a weak acid, the degree of dissociation is determined by the expression $\alpha = \sqrt{\frac{K_a}{C}}$, where $K_a$ is the acid dissociation constant and $C$ represents the initial molar concentration. Substituting the given values $K_a = 1.00 \times 10^{-5}$ and $C = 0.100$ mol/L into this equation yields $\alpha = \sqrt{\frac{1.00 \times 10^{-5}}{0.100}}$. Simplifying the term inside the square root results in $\sqrt{1.00 \times 10^{-4}}$, which equals $1.00 \times 10^{-2}$. Expressing this decimal value as a percentage by multiplying by 100 confirms that the extent of acid dissociation at equilibrium is 1.00%.

Determining the equilibrium constant for a composite reaction relies on the principle that if a reaction can be expressed as the sum of several elementary steps, the overall equilibrium constant is the product of the constants for those individual steps, adjusted for any stoichiometric coefficients or reversals. This approach, analogous to Hess's Law in thermochemistry, allows us to derive the relationship between the target reaction and the given equilibrium constants $K_1$, $K_2$, and $K_3$.

To obtain the oxidation reaction $2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O$, we first reverse the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$ to place $2NH_3$ on the reactant side, which changes its equilibrium constant to $\frac{1}{K_1}$. Next, we utilize the reaction $N_2 + O_2 \rightleftharpoons 2NO$ exactly as provided to obtain $2NO$ on the product side, contributing a factor of $K_2$. Finally, we take the reaction $H_2 + \frac{1}{2}O_2 \rightleftharpoons H_2O$ and multiply its coefficients by 3, which results in the production of $3H_2O$ and transforms its equilibrium constant to $K_3^3$.

Summing these three manipulated steps eliminates the intermediate species $N_2$ and $3H_2$ to yield the balanced target equation. Consequently, the equilibrium constant for the overall oxidation process is calculated by multiplying the modified constants together, resulting in the expression $\frac{K_2 K_3^3}{K_1}$

At 25°C, the relationship between the acidity and alkalinity of an aqueous solution is governed by the equation $pH + pOH = 14$. To determine the pOH, calculate the pH using the formula $pH = -\log[\text{H}_3\text{O}^{+}]$, which results in a value of 10 for a hydronium ion concentration of 1 x 10^-10 M. Subtracting this pH from 14 confirms that the pOH of the solution is 4.

The equilibrium constant expression is determined by the balanced chemical equation, requiring that the activities of products raised to their stoichiometric coefficients be divided by those of the reactants treated in the same manner. Because this system involves gaseous components, the constant $K_p$ must be expressed in terms of their partial pressures, and pure condensed phases such as $\mathrm{H}_2\mathrm{O}(\mathrm{l})$ are omitted from the expression entirely. Furthermore, the partial pressure of oxygen must be squared to account for its stoichiometric coefficient of two, a requirement that option C fails to meet. This option incorrectly attempts to use molar concentrations rather than partial pressures and omits the necessary exponential term for the oxygen gas, identifying it as the statement that does not hold true.

When dealing with extremely dilute acid solutions where the solute concentration is comparable to the auto-ionization of water, the hydrogen ions produced by water dissociation must be factored into the total concentration. For a $10^{-8}$ M HCl solution, the total hydrogen ion concentration is the sum of the acid contribution and the ions generated by water, represented as $x$, which makes the hydroxide ion concentration also equal to $x$. These values relate to the dissociation constant of water through the expression:

[H^+]$[OH^-]\$ = (10^{-8} + x)x = 10^{-14}

Solving this leads to the quadratic equation:

$x^2 + 10^{-8}x - 10^{-14} = 0$

Using the quadratic formula to solve for $x$ gives a concentration of approximately $0.9525 \times 10^{-7}$ M for the ions produced by water. Adding this to the $10^{-8}$ M contribution from the HCl results in a final total hydrogen ion concentration of $1.0525 \times 10^{-7}$ M.

A buffer solution resists significant changes in pH upon the addition of small amounts of acid or base by containing a weak acid and its conjugate base. Because nitrous acid ($HNO_2$) is a weak acid and sodium nitrite ($NaNO_2$) provides its corresponding conjugate base, this specific combination maintains the necessary equilibrium to qualify as an acidic buffer. Other combinations listed involve strong electrolytes that dissociate completely and fail to establish the necessary equilibrium to buffer against pH fluctuations.