This answer is correct because it is true that we use a catalyst to carry out Haber's process but it is the property of a catalyst that it does not alter the equilibrium position but provides an alternative path for the reaction.

E value of any cell at equilibrium is 0 but E⁰ is not zero. Further, E⁰ of a spontaneous cell is positive.

Numerically, b is equal to 4 times the volume occupied by 1 mole of gas molecules.

For a zero order reaction, A --> Products is inversely proportional to the rate constant.

Glucose is an aldohexose. Fructose is a ketohexose.

Total pressure = 5 atm. By Dalton's Law of partial pressure: P_{0xygen =} (5)*(0.2) = 1 atm. From Henry's Law:P_Oxygen = K_H * x1 = 2 * (n/n+10)Solving, n = 10

In both the isotopes, the number of protons is 1. Only the number of neutrons varies. If the reduced mass concept is not taken into account, then the radius would only depend on the central charge and not on the mass. Hence, these isotopes have the same atomic radii, although this is factually incorrect. Reduced mass concept holds in reality.

This option is incorrect because energy is given by , E_n = -13.6 / n² eV / atom . Hence, energy increases as we move away from the neucleus .

This option is incorrect because wave number is given by ; ϋ = RZ² [ 1 / n₁² - 1 / n₂² ]   . here , n₁ = 2   , n₂ = 3  &  Z = 1  . ϋ = R [ 1 / 4 - 1 / 9 ] = (5 / 36 ) R cm^-1 . hence, this option is correct .      

This option is correct because energy is given by ,  E_n = -13.6 / n² eV / atom . E_n -E_n-1 = -13.6 [ (1 / n² ) - (1 / (n-1)² ) ]  .  -1312 [ 1 /25 - 1] = 1259.52 kj / mol. .

An increase in pressure will favour the direction where there is lesser number of species.

This option is correct because equilibrium is always spontaneous. Being spontaneous means that the reaction proceeds with any external help or external driving force such as temperature of pressure.

This option is correct as at equilibrium the concentration of reactant and products are constant for a particular set of reactants and products and reaction quotient is equal to equilibrium constant of the reaction.

This option is correct because while the formula for equilibrium constant it is generalized as K = [products]/[reactants] ………………..where the concentration of pure reactants and products is taken to be unity.In the given reaction the concentration of reactant Zn(s) and product Cu(s) is to be taken unity and hence we will get the correct answer as K = [Zn²⁺]/[Cu²⁺]

The relation between K_p and K_c is given as K_p = K_c(RT)^∆ng………………∆ng is the difference between the number of moles of gaseous products and reactants.For this reaction ∆ng is 0. Hence K_p = K_c. Hence this is the correct answer.

This option is correct because I₂ is a solid and amount of solid present is not important as long as any equilibrium is present.

This option is correct as the given option may be written as K₃  =  K₁.K₂And when two reactions are added to give the third reaction then equilibrium constant of the third reaction is always the product of the equilibrium constant of the individual reactions. 

This option is correct and can be explained as follows: For the give n reaction the equilibrium constant is given byK = [Ba²⁺][SO₄^2-]……………………….or we can say that [Ba²⁺] and [SO₄^2-] are having inversely proportional relationship whose graph is given by a rectangular hyperbola.

The relation between K_p and K_c is given as K_p = K_c(RT)^∆ng………………∆ng is the difference between the number of moles of gaseous products and reactants.For the given reaction ∆ng is -2. Hence K_p = K_c(RT)^-2. Hence this is the correct answer.

This option is correct because for a reaction having enthalpy change zero is independent of change of temperature and favors neither formation of products or reactants.

This option is correct and can be explained as follows: The equilibrium constant for the reaction in the given question is given ask  =  [Butane]/[Isobutane].   However on decreasing the volume of the container the individual concentration of butane and isobutane increases but the overall change in equilibrium constant will be zero as the increase in concentrations of reactant and products will be cancelled.

This option is correct and can be explained as follows: Since ∆H0 for the reaction is +ve which means that reaction is endothermic in nature. Hence when temperature is increased then according to Le Chatelier's Principle it favors the forward reaction which nullifies the effect of increase in temperature hence, more nitric oxide is formed.

This answer is correct because all the reactants and products are in the aqueous phase hence, represent homogenous equilibria.

This option is correct and can be explained as follows: When NaCl(s) is added to the system it dissociates as NaCl(s)  ⇌  Na⁺(aq)  +  Cl^-(aq)The Cl^- ions obtained from the dissociation of solid NaCl will combine with Ag⁺ ions to precipitate AgCl. Hence the equilibrium will be shifted in forward direction in accordance with Le-Chatelier's principle and more Ag⁺ and SO₄^2- ions will be formed resulting in consumption of Ag₂SO₄ and hence its concentration will decrease.

This is the correct answer because the external effects on equilibrium were given by chemist Henry Louis Le Chatelier. Hence it is called Le Chatelier’s principle. According to Le Chatelier’s Principle when a system is subjected to any disturbance in its equilibrium state then equilibrium is shifted in the direction where the effect of the change in external condition is minimized.