Equilibrium state and Characteristics of K Questions in English

(B) For the reaction $N_2 + O_2 \rightleftharpoons 2NO$,chemical equilibrium is defined as the state where the rate of the forward reaction equals the rate of the backward reaction.
At this state,the concentrations of all reactants $(N_2, O_2)$ and products $(NO)$ remain constant over time.
Therefore,the correct statement is that the concentration of all substances is constant.

(C) Equilibrium is established in a chemical reaction when the rate of the forward reaction becomes equal to the rate of the backward reaction.

(B) For a physical or chemical process to reach equilibrium,the system must be closed so that no matter can escape or enter.
In option $(A)$,the vessel is open,so water vapor can escape.
In option $(B)$,the balloon is a closed system containing water and air,allowing for a dynamic equilibrium between liquid water and water vapor.
In option $(C)$,the vessel is open,so water vapor escapes continuously,preventing equilibrium.
Therefore,only option $(B)$ represents an equilibrium state.

(A) At equilibrium,the rate of the forward reaction is equal to the rate of the backward reaction.
This is the fundamental definition of chemical equilibrium.
When a reversible reaction is carried out in a closed vessel,a stage is reached where the forward and backward reactions proceed at the same speed.
This stage is known as chemical equilibrium.

(C) In a chemical reaction $A \rightleftharpoons B$,equilibrium is defined as the state where the rate of the forward reaction $(A \rightarrow B)$ is equal to the rate of the backward reaction $(B \rightarrow A)$.
Therefore,the correct option is $(C)$.

(B) If the rate of forward reaction equals the rate of backward reaction,then the reaction is said to be in equilibrium. At this stage,there is no further change in the concentrations of reactants as well as products.

(D) . At equilibrium,the rate of the forward reaction is equal to the rate of the backward reaction.
This means that the rate at which $NH_3$ is produced is equal to the rate at which $NH_3$ decomposes back into $N_2$ and $H_2$.

(B) At equilibrium,the rate of the forward reaction is equal to the rate of the reverse reaction. Therefore,the correct option is $B$.

(C) Chemical equilibrium is considered dynamic because,at equilibrium,the rate of the forward reaction equals the rate of the backward reaction.
Even though the concentrations of reactants and products remain constant over time,the reactions do not stop; they continue to occur at equal rates in both directions.

(A) The extent of a reaction is determined by the magnitude of the equilibrium constant $(K)$.
If $K > 10^3$,the reaction proceeds towards completion,meaning the concentration of products is much higher than that of reactants.
Among the given options,$K = 10^3$ represents the largest value,indicating that the reaction proceeds furthest towards completion.

(D) The equilibrium constant $(K_c)$ is a characteristic property of a specific chemical reaction at a constant temperature.
It depends only on the temperature and is independent of the initial concentrations of the reactants or products.
Therefore,changing the concentrations of the reactants does not affect the value of $K_c$.

(C) The equilibrium constant ($K_c$ or $K_p$) is a constant value for a given reaction at a specific temperature.
It is independent of the initial concentrations of the reactants or products.
It only changes when the temperature of the system is changed.

(A) The dissociation reaction of ammonia is: $2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$.
$K_p$ is the equilibrium constant,which depends only on temperature for a given reaction.
Therefore,$K_p$ does not change with pressure at a constant temperature.
Thus,option $A$ is correct.

(C) Chemical equilibrium is a state in a reversible reaction where the rate of the forward reaction equals the rate of the backward reaction. At this point,the concentrations of reactants and products remain constant over time. Therefore,the correct option is $(C)$.

(C) The equilibrium constant $(K_c)$ for a given reaction depends only on the temperature.
Since the temperature remains constant at $721 \ K$,the value of the equilibrium constant will remain unchanged regardless of the addition of reactants or products.
Therefore,the value of the equilibrium constant remains $50$.

(D) The equilibrium constant ($K_c$ or $K_p$) for a given chemical reaction is a function of temperature only.
It is independent of the initial concentrations of reactants,the total pressure of the system,the volume of the reaction vessel,or the presence of a catalyst.
The relationship between the equilibrium constant and temperature is given by the van't Hoff equation: $\ln \frac{K_2}{K_1} = \frac{\Delta H^\circ}{R} (\frac{1}{T_1} - \frac{1}{T_2})$.
Therefore,the value of the equilibrium constant depends on the temperature.

(C) The equilibrium constant $K_c$ is a function of temperature only.
Since the temperature is kept constant at $25\,^oC$,changing the concentration of the reactants or products will shift the equilibrium position according to Le Chatelier's Principle,but it will not alter the value of the equilibrium constant $K_c$.
Therefore,$K_c$ remains the same.

(C) The equilibrium constant $(K_c)$ of a reaction depends only on the temperature.
It is independent of the initial concentrations,pressure,or the volume of the container.
Since the temperature remains constant,the value of the equilibrium constant will remain $64$.

(D) The equilibrium constant $(K_c)$ is a function of temperature only.
It remains unaffected by changes in concentration,pressure,or the addition of a catalyst.
Therefore,when the mixture is heated,the value of the equilibrium constant changes.

(D) For any chemical reaction,the value of the equilibrium constant $(K)$ is a function of temperature. It changes with temperature for both exothermic and endothermic reactions. However,the addition of a catalyst does not affect the equilibrium constant because it increases the rate of both forward and backward reactions equally.

(B) The equilibrium constant $(K_c)$ is a characteristic property of a reaction at a given temperature.
It depends only on the temperature and is independent of the initial concentrations,pressure,or volume of the reaction container.
Therefore,changing the volume of the container does not change the value of $K_c$.
Thus,the value remains $64$.

(D) The equilibrium constant ($K_c$ or $K_p$) for a given reversible reaction depends only on the temperature.
It is independent of the volume of the reaction vessel,pressure,or the initial concentration of reactants.
For the reaction $H_2(g) + Cl_2(g) \rightleftharpoons 2HCl(g)$,the change in the number of moles of gaseous products and reactants is $\Delta n_g = 2 - (1 + 1) = 0$.
Since the equilibrium constant is independent of volume,it will remain the same regardless of the flask size.

(D) The equilibrium constant ($K_c$ or $K_p$) is a function of temperature only for a given reaction.
It does not change with the addition of a catalyst,changes in pressure,or changes in the initial concentrations of reactants or products.
Therefore,the correct option is $(D)$.

(B) For the phase transition $H_2O_{(s)} \rightleftharpoons H_2O_{(l)}$ at $0\,^\circ C$ $(273.15\,K)$ and $1\,atm$ pressure,the system is in a state of equilibrium.
At equilibrium,the change in Gibbs free energy is $\Delta G = 0$.
From the thermodynamic relation $\Delta G = \Delta H - T\Delta S$,substituting $\Delta G = 0$ gives $0 = \Delta H - T\Delta S$.
Therefore,$\Delta H = T\Delta S$.

(A) The extent of a reaction is determined by the magnitude of the equilibrium constant $(K)$.
$A$ Higher the value of the equilibrium constant $(K)$,the greater the concentration of products relative to reactants at equilibrium.
$B$ Therefore,a larger value of $K$ indicates that the reaction proceeds further toward completion.
$C$ Comparing the given values: $10^3 > 10 > 1 > 10^{-2}$.
$D$ Thus,$K = 10^3$ represents the case where the reaction goes farthest to completion.

(C) catalyst provides an alternative pathway with a lower activation energy for both the forward and backward reactions.
Consequently,it increases the rate of both the forward and reverse reactions to the same extent.
This allows the system to reach the state of chemical equilibrium faster without changing the equilibrium constant or the equilibrium concentrations of reactants and products.

(C) catalyst is a substance that increases the rate of both the forward and backward reactions to the same extent by lowering the activation energy.
Consequently,it helps the system reach the equilibrium state in a shorter amount of time without altering the position of the equilibrium or the enthalpy change of the reaction.
Therefore,the correct option is $C$.

(B) catalyst provides an alternative reaction pathway with a lower activation energy. It increases the rate of both the forward and backward reactions to the same extent. Therefore,it does not change the position of the equilibrium or the equilibrium concentrations of the reactants and products. The correct option is $(B)$.

(C) catalyst provides an alternative reaction pathway with a lower activation energy,which increases the rate of both the forward and backward reactions equally. Therefore,it does not change the equilibrium constant $(K_{eq})$ of the reaction.

(B) At equilibrium,the Gibbs free energy change of the system is zero.
Therefore,for the process $H_2O_{(\ell)} \rightleftharpoons H_2O_{(g)}$ at its boiling point ($373 \, K$ and $1 \, \text{atm}$),the condition is $\Delta G = 0$.

(B) At $0^{\circ}C$ and $1 \ atm$ pressure,ice and water are in equilibrium.
For a process at equilibrium,the change in Gibbs free energy is $\Delta G = 0$.
We know the relation $\Delta G = \Delta H - T\Delta S$.
Substituting $\Delta G = 0$,we get $0 = \Delta H - T\Delta S$,which implies $\Delta H = T\Delta S$.

(C) Chemical equilibrium is a state in a reversible reaction where the rate of the forward reaction becomes equal to the rate of the backward reaction. At this point,the concentrations of reactants and products no longer change with time,meaning they become constant. Therefore,the most accurate definition of the establishment of equilibrium is when the rate of the forward reaction equals the rate of the backward reaction.

(A) $1$. The equilibrium constant $(K)$ is a function of temperature only for a given reaction. It changes when the temperature changes.
$2$. It is independent of the initial concentrations of the reactants or products.
$3$. For the reaction $3Fe_{(s)} + 4H_2O_{(g)} \rightleftharpoons Fe_3O_{4_{(s)}} + 4H_{2_{(g)}}$,the equilibrium constant expression is $K_c = \frac{[H_2]^4}{[H_2O]^4}$. This reaction involves gases and solids. In an open vessel,the gaseous products escape,preventing equilibrium from being established. Therefore,the equilibrium constant is only defined for a closed system.
$4$. If a reaction is multiplied by a factor $n$,the new equilibrium constant becomes $K^n$. Thus,if multiplied by $2$,it becomes $K^2$,not $2K$.

(C) The equilibrium constant $(K_c)$ for a given reaction depends only on the temperature.
Since the temperature is kept constant,the equilibrium constant will remain unchanged regardless of the change in concentration of the reactants or products.

(D) The equilibrium constant ($K_c$ or $K_p$) is a function of temperature only for a given reaction. It does not change with the addition of a catalyst,changes in pressure,or changes in the initial concentrations of reactants or products.

(B) catalyst provides an alternative reaction pathway with a lower activation energy.
It increases the rate of both the forward and backward reactions to the same extent.
Therefore,it does not affect the position of the equilibrium or the equilibrium constant $(K_{eq})$.

(C) The equilibrium constant $(K_c)$ of a chemical reaction depends only on the temperature of the system.
It is independent of the presence of a catalyst.
$A$ catalyst only increases the rate of both the forward and backward reactions equally,thereby allowing the system to reach equilibrium faster without changing the position of the equilibrium or the value of the equilibrium constant.
Therefore,the equilibrium constant remains $4 \times 10^4$.

(B) The equilibrium constant ($K_c$ or $K_p$) is a characteristic property of a reversible reaction at a given temperature.
It depends only on the temperature and is independent of the initial concentrations of reactants or products,pressure,or the presence of a catalyst.
Therefore,changing the concentration of the reactant does not change the value of the equilibrium constant.

(C) The equilibrium constant ($K_p$ or $K_c$) of a chemical reaction depends only on the temperature of the system.
Since the temperature is kept constant,the equilibrium constant will remain unchanged.
Adding an inert gas like $He$ at constant pressure causes the volume of the system to increase,which may shift the equilibrium position,but it does not affect the value of the equilibrium constant itself.

(C) The equilibrium constant $(K_c)$ for a given reaction depends only on the temperature of the system.
It is independent of the initial concentrations,pressure,or the volume of the reaction container.
Since the temperature remains constant,changing the volume of the container does not affect the value of the equilibrium constant.
Therefore,the equilibrium constant remains $64$.

(D) The equilibrium constant $(K_c)$ for a reversible reaction is a function of temperature only.
It does not depend on the initial concentrations of the reactants or products,pressure,or the presence of a catalyst.
Therefore,if the concentration of the reactants is doubled at a constant temperature,the equilibrium constant $(K_c)$ will remain constant.

(B) The Gibbs free energy change $(\Delta G)$ determines the spontaneity of a reaction:
$1$. If $\Delta G < 0$,the reaction is spontaneous.
$2$. If $\Delta G > 0$,the reaction is non-spontaneous.
$3$. If $\Delta G = 0$,the reaction is at equilibrium.

(A) For the dissociation of ammonia: $2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$.
At constant temperature,the equilibrium constant ($K_p$ or $K_c$) depends only on temperature and remains constant regardless of changes in pressure or concentration.
Therefore,$K_p$ does not change with pressure.

(D) Chemical equilibrium is a state in a reversible reaction where the rate of the forward reaction equals the rate of the backward reaction.
For the reaction $x \rightleftharpoons y$,equilibrium is attained when the rate of conversion of $x$ to $y$ (forward rate) is equal to the rate of conversion of $y$ to $x$ (backward rate).
At this point,the concentrations of reactants and products remain constant over time.

(B) The equilibrium constant $(K)$ for a given chemical reaction depends only on the temperature of the system.
It is independent of the initial concentrations of the reactants or products.
Since the temperature remains constant in both experiments,the value of the equilibrium constant will remain the same,i.e.,$K$.

(C) In a reversible chemical reaction,when the rate of the forward reaction becomes equal to the rate of the backward reaction,the concentrations of reactants and products remain constant with time. This state is defined as chemical equilibrium.

(A) The stability of a compound is inversely proportional to the equilibrium constant $(K)$ of its decomposition reaction.
$A$ smaller value of $K$ indicates that the reaction proceeds less towards the product side,meaning the reactant is more stable.
Comparing the given equilibrium constants:
$K_A = 6.7 \times 10^{16}$
$K_B = 2.2 \times 10^{30}$
$K_C = 1.2 \times 10^{34}$
$K_D = 3.5 \times 10^{33}$
Since $K_A$ is the smallest,the oxide $NO_2$ is the most stable among the given options.

(D) The equilibrium constant $(K_c)$ for a reversible reaction is a function of temperature only. It does not depend on the initial concentrations of the reactants or products,the presence of a catalyst,or the pressure of the system. Therefore,if the concentration of the reactants is doubled,the equilibrium constant $(K_c)$ will remain the same.