NEET Chemistry Questions: Equilibrium

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Three gaseous equilibria have values of equilibrium constants as $k_1 , k_2, k_3 $ . resp. (i) $ A +B \rightleftharpoons C $ (ii) $ B+C \rightleftharpoons P+Q $ (iii) $ A+2B \rightleftharpoons P+Q $. What is the relation between $k_1 ,k_2, k_3$ . (?)




Addition of (1) and (2) gives (3) then $ K_1 \times K_2 = { [C] \over [A] [ B] ] } \times { [P] [Q] \over [ B] [C] } = { [P] [Q] \over [ A] [ B ] ^2 } = K_3 $ $ \therefore K_3 = K_1 \times K_2 $ When the addition of equilibria leads to another equilibria then the product of their equilibria constants gives the equilibria constant of the resultant equilibrium.
Three gaseous equilibria have value of equilibrium constants as $K_1, K_2 , K_3$ respectively. (i)$ N_2 + O_2 \rightleftharpoons 2 NO (K_1) $ (ii) $ N_2 + 2O_2 \rightleftharpoons 2 NO_2 (K_2 ) $ (iii) $ 2NO +2 O_2 \rightleftharpoons 2 NO_2 (K_3 ) $ What is the relation between $ K_1, K_2 and K_3 (?) $




The given equilibria can be analyzed to find the relationship between the equilibrium constants. Rearranging and combining the given reactions, we get the relation $ K_2 = K_1 imes K_3 $. This comes from the fact that if you add the first equilibrium (i) and the third equilibrium (iii), you get the second equilibrium (ii), and the equilibrium constant for the added reactions is the product of their individual constants.
The equilibrium constant for the reaction, $ N_{2(g)} + O_{2(g) } \rightleftharpoons 2 NO_{(g)} is 4.4 \times 10 ^ {-4 } $ at 2000k temp In presence ofa catalyst, equilibrium is attained ten times faster. Therefore the equilibrium constant, in presence of catalyst at 2000K is........




The value of equilibrium constant does not change in presence of catalyst.
For the following reaction in gaseous phase $ CO + 1/2 O_2 \rightleftharpoons CO_2 $ $ KC/ KP $ is




Formula: $ Kp = Kc(RT) ^ {\triangle n_{ (g)} } $ $ \triangle_{(g)} = n_p - n_r $
For the reaction equilibrium, $ N_2O_{4(g )} \rightleftharpoons 2NO_{( g)} the concentration of N_2O_4 and NO_2 at equilibrium are 4.8 \times 10 ^ {-2} and 1.2 \times 10 ^ {-2} molL^{-1} respectively. The value of K_C $ for the reaction is :




For the given reaction equilibrium $N_2O_{4(g)} ightleftharpoons 2NO_{2(g)}$, the equilibrium constant $K_C$ is calculated using the formula: $$ K_C = rac{[ NO_2]^2}{[ N_2O_4]} $$ Given the concentrations: $$ [NO_2] = 1.2 imes 10^{-2} ext{ mol L}^{-1} $$ $$ [N_2O_4] = 4.8 imes 10^{-2} ext{ mol L}^{-1} $$ Substitute these values into the formula: $$ K_C = rac{(1.2 imes 10^{-2})^2}{4.8 imes 10^{-2}} = rac{1.44 imes 10^{-4}}{4.8 imes 10^{-2}} = 3 imes 10^{-3} ext{ mol L}^{-1} $$ Therefore, the correct answer is $3 imes 10^{-3} ext{ mol L}^{-1}$.

Which one of the following statements is not true ?





$ H_2O is also present in 1 \times 10^{-8} M HCl solution So due to self ionisation of H_2O $ , $ H_2 O \rightleftharpoons H^+ + OH^- $ $ 1 \times 10 ^ {-7} $ M at 298 K

What is the equilibrium expression for the reaction. $ P_{4(s)} + 5O_{2(g)} \rightleftharpoons P_4O_{10(s)} $




For the reaction $P_{4(s)} + 5O_{2(g)} ightleftharpoons P_4O_{10(s)}$, the equilibrium constant expression $K_C$ involves only the concentrations of gases and aqueous species. Solids are not included in the equilibrium expression. Thus, the expression for $K_C$ is: $$ K_C = rac{1}{[O_2]^5} $$ Therefore, the correct answer is $K_C = rac{1}{[O_2]^5}$.

For the reaction CO(g) + Cl2(g) ----> COCl2(g) then  Kp/ Kc  is equal to





No explanation available.
The equilibrium constant for the reaction $ N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)} at temperature. 300 k is 4 \times 10^{-4 } $.The value of Kc for the reaction $ NO_{(g)} \rightleftharpoons {1 /2 } N_{2(g) } + { 1/2} O_{2(g) } $ at the same temperature is




$ N_{2 (g)} + O_{ 2 (g) } \rightleftharpoons 2 NO_{(g)} $ $ K_p = { P^2 NO \over P_ {N_2} \times P_{O_2} $ For reaction $ NO(g) \rightleftharpoons {1 \over 2} O_{2 (g) } + { 1 \over 2 } N_2 $ $ K_p = { P ^ {1/2} _{O_2} \times P^ {NO} = { 1 \over ( Kp ) ^ {1/2} }= { 1 \over ( 4 \times 10 ^ {-4} ) ^ {1/2} } = 50 $
Hydrogen ion concentration in mol/ L in a solution of $ P^H = 5.4 $ will be,




$ P ^ H = 5.4 $ $ P ^ H = - log [ H^+] $ $ 5.4 = - - log [ H ^ + ] $ $ 6.6 = 3.981 \times 10 ^ {-6} = [ H^ + ] $
$ In a reaction, CO_{(g)} + 2H_{2(g)} \rightleftharpoons CH_{(g)} OH \triangle H^ \circ = -92 KJ mol ^ {-1 } $ concentration of hydrogen,carbon monoxide and methanol become constant at equilibrium. what willhappen if an inert gas is added to the system.?




When an inert gas is added to a system at constant volume, the total pressure increases but the partial pressures of the reacting gases remain the same. Therefore, the equilibrium state remains undisturbed as the concentrations of the reactants and products do not change.
At 473 k, equilibrium constant $ K_C for the reaction PCl_{5(s)} \rightleftharpoons PCl_{3(g)}+ P Cl_{2(g)} is 8.3 \times 10 ^ {-3} what is the value of K_C $ for the reverse reaction at the same temperature ?




The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction. Hence, if $ K_C $ for the forward reaction is $ 8.3 imes 10^{-3} $, then $ K_C $ for the reverse reaction is $ \frac{1}{8.3 imes 10^{-3}} = 120.48 $.
At 473 k, equilibrium constant KC for a reaction, $ PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)} is 8.3 \times 10 ^ {-3 } ( \triangle H = 124 Kjmol ^{-1 }) $ What would be the effect on $ K_C $ if (i) more PCls is added (ii) the pressure is increased (iii) the temp is increased




For the reaction $ PCl_{5(g)} ightleftharpoons PCl_{3(g)} + Cl_{2(g)} $, the equilibrium constant $ K_C $ is only affected by temperature. Therefore, if more $ PCl_5 $ is added or if the pressure is increased, $ K_C $ remains unchanged. However, since the reaction is endothermic (ΔH > 0), increasing the temperature will increase $ K_C $.
The equilibrium constant for the reaction $ N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)} is 4 \times 10 ^ {-4} at 2000K temperature. what is the value of kc for the reaction 3/2 N_{2(g)} + 3/2 O_{2 (g)} \rightleftharpoons 3 NO_{(g)} $




To find the equilibrium constant for the reaction $ \frac{3}{2} N_{2(g)} + \frac{3}{2} O_{2(g)} \rightleftharpoons 3 NO_{(g)} $, we need to relate this reaction to the given reaction $ N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)} $. The given equilibrium constant for $ N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)} $ is $ K_c = 4 \times 10^{-4} $. If we multiply the coefficients of the given reaction by $ \frac{3}{2} $, we get $ \frac{3}{2} N_{2(g)} + \frac{3}{2} O_{2(g)} \rightleftharpoons 3 NO_{(g)} $. The new equilibrium constant is related to the original one by raising it to the power of $ \frac{3}{2} $. Therefore, $ K_c' = (4 \times 10^{-4})^{\frac{3}{2}} = 8 \times 10^{-6} $.
For the reversible reaction $ N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} at 500 \circ c the value of K_P is 1.44 \times 10 ^{-5} $ . what would be the value of KC for the same reaction




No explanation available.