Equilibrium state and Characteristics of K Questions in English

(C) catalyst increases the rate of both forward and backward reactions equally by lowering the activation energy.
It does not affect the position of equilibrium or the equilibrium constant $(K_c)$.
Therefore,the statement that the value of $K_c$ decreases in the presence of a catalyst is incorrect.

(C) At equilibrium,the rate of the forward reaction is equal to the rate of the backward reaction.
Therefore,the rate of the backward reaction at equilibrium is $0.02 \ mol \ L^{-1} \ min^{-1}$.

(B) For a general reaction,the equilibrium constant is defined as $K_c = \frac{[\text{Product}]}{[\text{Reactant}]}$.
If $[\text{Product}] > [\text{Reactant}]$,then the ratio $K_c$ must be greater than $1$.
Comparing the given values:
$A: K = 0.001 < 1$
$B: K = 10 > 1$
$C: K = 0.005 < 1$
$D: K = 0.01 < 1$
Therefore,in reaction $B$,the concentration of the product is higher than the concentration of the reactant.

(C) In the equilibrium $H_{2} + I_{2} \rightleftharpoons 2 HI$,if at a given temperature,the concentrations of the reactants are increased,the value of the equilibrium constant $K_{C}$ will remain the same.
This is because the equilibrium constant $K_{C}$ is a function of temperature only and does not depend on the initial molar concentrations of the reactants or products.
The correct option is $C$.

(C) The equilibrium constant ($K_c$ or $K_p$) is a characteristic value for a given reaction at a specific temperature. It depends only on the temperature and is independent of the initial concentrations of the reactants or products,pressure,or the presence of a catalyst. Therefore,if the concentration of any reactant is doubled,the equilibrium constant remains unchanged.

(A) catalyst provides an alternative reaction pathway with a lower activation energy,which increases the rate of both the forward and backward reactions equally.
Since the rate of both reactions increases by the same factor,the equilibrium position remains unchanged.
The equilibrium constant $(K_{eq})$ is a function of temperature only for a given reaction.
Since the temperature remains constant at $2000 \ K$,the equilibrium constant remains $4 \times 10^{-4}$.

(D) For the given equilibrium $H_{2}O_{(l)} \rightleftharpoons H_{2}O_{(g)}:$
$(I)$ At equilibrium,the Gibbs free energy change is $\Delta G = 0$.
$(II)$ Entropy change,$\Delta S$,measures the randomness of the system. Since the vapor state is more disordered than the liquid state,the entropy increases,meaning $\Delta S > 0$.
$(III)$ The conversion of liquid $H_{2}O$ to vapor $H_{2}O$ is an endothermic process,which requires the absorption of heat energy. Therefore,the enthalpy change is $\Delta H > 0$.
Thus,the correct set of conditions is $\Delta G = 0, \Delta H > 0, \text{ and } \Delta S > 0$.
Therefore,option $(D)$ is the correct answer.