Equilibrium MCQs for NEET — Chemistry Questions with Answers

$ H_{2(g) } + CO_{2(g)} \rightleftharpoons CO_{(g)} + H_2 O_{(g)}$ initial

conc. 1 1 0 0 At eqm. (1-x) (1-x) x x $ Kp = { P_{CO} \times P_{H_2O} \over P_{H_2} \times P_{CO_2} } ={ x^2 \over (1-x)^2 $

For the given reaction \( CH_{4(g)} + 2 O_{2(g)} \rightleftharpoons CO_{2(g)} + 2 H_2O_{(L)} \Delta H = -170.8 \, KJ \, mol^{-1} \), we need to determine which statement is not true. The equilibrium constant expression for a reaction involving gases is typically given in terms of partial pressures or concentrations of the gaseous reactants and products. However, in this case, \( H_2O \) is in the liquid state, so its concentration is taken as constant and omitted from the equilibrium expression. Thus, the equilibrium constant expression should be: \[ Kp = \frac{[CO_2]}{[CH_4][O_2]^2} \] Therefore, option o4 is not true.

The reaction quotient (Q) gives the ratio of the concentrations of products to reactants at any point in time. For the reaction \( N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_3 \), \[ Q = \frac{[NH_3]^2}{[N_2][H_2]^3} \] The direction in which the reaction will proceed depends on the comparison between Q and the equilibrium constant \( K_c \). If \( Q > K_c \), the reaction will proceed in the reverse direction (from right to left) to reach equilibrium. Therefore, the correct option is \( Q > K_c \).

The formation of $SO_3$ from $SO_2$ and $O_2$ is an exothermic reaction, as indicated by the negative enthalpy change ($ riangle H = -45.0 ext{ Kcal}$). According to Le Chatelier's Principle, increasing the temperature would shift the equilibrium to the left, favoring the reactants and thus not favoring the formation of $SO_3$. Therefore, high temperature is not favorable for $SO_3$ formation.

The most important buffer system in blood consists of carbonic acid ($H_2CO_3$) and bicarbonate ($HCO_3^-$). This buffer system maintains the pH of blood by neutralizing acids and bases, ensuring that the pH remains within the narrow range necessary for proper physiological function.

The strength of an acid is inversely related to its pKa value; the lower the pKa value, the stronger the acid. Among the given options, a pKa value of 1 is the lowest, indicating that it corresponds to the strongest acid.

The ionization constant of hydrofluoric acid (HF) can be found using the relationship between the ionization constant of the acid (Ka) and the base dissociation constant (Kb) of its conjugate base (fluoride ion, F^-). The relationship is given by: $$ K_a imes K_b = K_w $$ where $$ K_w $$ is the ion-product constant for water at 25°C ($1 imes 10^{-14}$). Given $$ pK_b = 10.83 $$, we first find $$ K_b $$: $$ K_b = 10^{-pK_b} = 10^{-10.83} = 1.48 imes 10^{-11} $$ Then, $$ K_a = rac{K_w}{K_b} = rac{1 imes 10^{-14}}{1.48 imes 10^{-11}} = 6.75 imes 10^{-4} $$ Hence, the correct answer is $ 6.75 imes 10 ^ {-4} $.