Equilibrium practice · Mixed difficulty

The correct answer is A ($\frac{1}{{K_1^2}} + \frac{1}{{{K_2}}}$) because it represents the equilibrium constant for the reverse of the given reactions to form N2O4 from N2 and O2.

Option D ($2.7 \times {10^{ - 2}}\;M$) is correct because it is the hydrogen ion concentration calculated at the equivalence point after the weak base is neutralized by the strong acid.

Option B ($\frac{1}{1+\sqrt{2}}$) is the correct answer because it represents the degree of association calculated from the equilibrium constant.

Option D (0.22) is correct because it is calculated by setting up an ICE table and solving for the degree of dissociation using the equilibrium constant expression.

Option B ($1 \times {10^{ - 2}}\;M$) is correct because pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration.

Option B ($w(N_2) + w(H_2) + w(NH_3) = 102$ g at $t=2t_1$) is correct because at equilibrium, the total mass remains constant, and 34g of ammonia is formed from N2 and H2.

Option D (If the assertion and reason both are false) is correct. The assertion is false because $K_p = K_c$ only when $\Delta n_g = 0$, and the reason is false because the concentration of gaseous reactant is not always unity.

Option C (1.32 g) is correct because it represents the mass of 95% pure CaCO3 required to neutralize the given HCl solution, calculated using stoichiometry.

Option A (The value of equilibrium constant is changed in the presence of catalyst in the reaction at equilibrium) is correct because a catalyst only speeds up the rate at which equilibrium is reached but does not change the value of the equilibrium constant.

Option C ($10^{-7}$) is correct because it represents the approximate ratio of [H3A]/[A3-] at the second equivalence point, calculated using the given Ka values.

Option A ($\mathrm{CaO} < CuO < {\mathrm{H}_2}\mathrm{O} < C{\mathrm{O}_2}$) is the correct order because it reflects the increasing non-metallic character and acidity.

Option D is correct because the difference in pH at 1/4 and 3/4 neutralization stages of a weak acid is given by 2log3 based on the Henderson-Hasselbalch equation.

Option A ($1.0\mathrm{ }\;atm$) is correct because it is the value of Kp calculated using partial pressures at equilibrium.

Option A (Phenolphthalein) is correct because phenolphthalein changes color in the pH range of 8.3 to 10.0, which includes the range of 8 - 9.8.

Option A (Dissociation of $PC{l_5}$ increases) is correct because adding an inert gas at constant pressure increases the volume, favoring the side with more moles of gas, which is the dissociated side.

The correct answer is C (${K_2}/{K_1}$) because it represents the equilibrium constant obtained by manipulating the given reactions to match the target reaction.

Option B (Zero) is correct because a reaction attains equilibrium when the Gibbs energy change (ΔG) is zero, indicating that the system is at its lowest energy state.

Option D (Sodium acetate) is correct because sodium acetate is a salt of a strong base (NaOH) and a weak acid (acetic acid), resulting in a basic solution due to hydrolysis of the acetate ion.

Option A ($1 \times {10^{ - 3}}$) is correct because the degree of hydrolysis can be calculated using the formula h = sqrt(Kh/C) where Kh = Kw/Ka.

Option A (${\left( {Conc} \right)^1}$) is correct because the unit of equilibrium constant K is determined by the change in the number of moles in the reaction.

Option C ($\mathrm{PC}{\mathrm{l}_5}$ is less) is correct because adding CO shifts the second equilibrium to the left, decreasing Cl2, which then shifts the first equilibrium to the left, decreasing PCl5.

Option B ($250\,\,\mathrm{K}$) is correct because at equilibrium, $\Delta G = 0$, and we can solve for the temperature using the given $\Delta H$ and $\Delta S$ values.

Option D ($\mathrm{N}{\mathrm{H}_3}$) is correct because ammonia has a lone pair of electrons that it can donate, making it a Lewis base.

Option A (4.67, 9.61) is correct because it is calculated using the Henderson-Hasselbalch equation for polyprotic acids.

Option C (Solubility product) is correct because salting out involves decreasing the solubility of soap by adding a common ion, which is related to the solubility product principle.

Option B (8) is correct because in an amphoteric buffer where [H+] = [HX], pH = pKa.

Option A is correct because the solubility product is related to solubility by the equation $L_{{S}}=S^{p+q} \cdot p^p \cdot q^q$ for a sparingly soluble salt ${A}_{{p}} {B}_{{q}}$.

Option A ($K=e^{-\Delta G^{\circ} / R T}$) is correct because it correctly expresses the equilibrium constant (K) in terms of the standard Gibbs free energy change (ΔG°), the gas constant (R), and the temperature (T).

Option C is correct because at equilibrium, the rate of the forward reaction equals the rate of the backward reaction.

Option C ($S{n^{4 + }}$) is correct because $S{n^{2 + }}$ loses two electrons to get oxidized to $S{n^{4 + }}$ when ${K_2}C{r_2}{O_7}$ is added.