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The reaction which is in dynamic equilib...

The reaction which is in dynamic equilibrium, ensured us, that the reaction is reversible . But if that the reaction is in equilibrium. The reaction quotient predict either the reversible reaction is in equilibrium or tries to achieve equilibrium. In those reactions which have not achieved equilibrium, we obtain reaction quotient `Q_(c)` in place of equilibrium constant `(K_(c))` by substituting the concentration of reactant and product at the time, at whih we have to calculate the value of `Q_(c)` . To determine the direction at which the net reaction will proceed to achieve equilibrium, we compare values of `Q_(c)` and `K_(c)`. The three possible cases are shown as comparison of `K_(c)` and `Q_(c)` in the following figures.

Change in Gibbs free energy, i.e., `Delta G` is the driving force of any reaction.
For spontaneous reaction , `Delta G =-ve`
For non-spontaneous reaction , `Delta G=+ve`
For reaction at equilibrium , `Delta G =0`
Thermodynamically, we know that
`Delta G= Delta G^(@)+ RT ln Q`, where `Q` is reaction quotient and `Delta G^(@)=` change in Gibbs energy at standard condition.
For equilibrium `A(g) hArr B(g) (K_(eq) =1.732)` If the pressure of the system [varied by introducing a stream of `A (g)` and B (g) is represented by the curve at constant temperature T.

What will be the value of difference of standard Gibbs free energy to Gibbs free energy change at point Q in the figure above ?

A

`-ve`

B

`+ve`

C

0

D

cannot be predicted

Text Solution

Verified by Experts

The correct Answer is:
A
`Delta G=DeltaG^(@)+RT ln Q`
`:. Delta G^(@)-Delta G=-RT ln Q`
`=- ve (" Q gt 1.732 )`
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At equilibrium the reaction quotient is greater than equilibrium constant.

The reaction which is in dynamic equilibrium, ensured us, that the reaction is reversible . But if that the reaction is in equilibrium. The reaction quotient predict either the reversible reaction is in equilibrium or tries to achieve equilibrium. In those reactions which have not achieved equilibrium, we obtain reaction quotient Q_(c) in place of equilibrium constant (K_(c)) by substituting the concentration of reactant and product at the time, at whih we have to calculate the value of Q_(c) . To determine the direction at which the net reaction will proceed to achieve equilibrium, we compare values of Q_(c) and K_(c) . The three possible cases are shown as comparison of K_(c) and Q_(c) in the following figures. Change in Gibbs free energy, i.e., Delta G is the driving force of any reaction. For spontaneous reaction , Delta G =-ve For non-spontaneous reaction , Delta G=+ve For reaction at equilibrium , Delta G =0 Thermodynamically, we know that Delta G= Delta G^(@)+ RT ln Q , where Q is reaction quotient and Delta G^(@)= change in Gibbs energy at standard condition. For equilibrium A(g) hArr B(g) (K_(eq) =1.732) If the pressure of the system [varied by introducing a stream of A (g) and B (g) is represented by the curve at constant temperature T. Suppose the equilibrium system N_(2)O_(4)(g) hArr 2NO_(2)(g) N_(2)O_(4)(g) is in a cylinder fitted with a movable piston . Which of the following statements is correct ?

Knowledge Check

  • The reaction which is in dynamic equilibrium, ensured us, that the reaction is reversible . But if that the reaction is in equilibrium. The reaction quotient predict either the reversible reaction is in equilibrium or tries to achieve equilibrium. In those reactions which have not achieved equilibrium, we obtain reaction quotient Q_(c) in place of equilibrium constant (K_(c)) by substituting the concentration of reactant and product at the time, at whih we have to calculate the value of Q_(c) . To determine the direction at which the net reaction will proceed to achieve equilibrium, we compare values of Q_(c) and K_(c) . The three possible cases are shown as comparison of K_(c) and Q_(c) in the following figures. Change in Gibbs free energy, i.e., Delta G is the driving force of any reaction. For spontaneous reaction , Delta G =-ve For non-spontaneous reaction , Delta G=+ve For reaction at equilibrium , Delta G =0 Thermodynamically, we know that Delta G= Delta G^(@)+ RT ln Q , where Q is reaction quotient and Delta G^(@)= change in Gibbs energy at standard condition. For equilibrium A(g) hArr B(g) (K_(eq) =1.732) If the pressure of the system [varied by introducing a stream of A (g) and B (g) is represented by the curve at constant temperature T. Which of the following is the correct statement at point R in the figure above ?

    A
    The reaction will move in backward direction in order to gain equilibrium
    B
    The reaction will move in forward direction in order to gain equilibrium
    C
    `Delta G` of the reaction will be zero
    D
    Cannotbe predicted.

  • The reaction which is in dynamic equilibrium, ensured us, that the reaction is reversible . But if that the reaction is in equilibrium. The reaction quotient predict either the reversible reaction is in equilibrium or tries to achieve equilibrium. In those reactions which have not achieved equilibrium, we obtain reaction quotient Q_(c) in place of equilibrium constant (K_(c)) by substituting the concentration of reactant and product at the time, at whih we have to calculate the value of Q_(c) . To determine the direction at which the net reaction will proceed to achieve equilibrium, we compare values of Q_(c) and K_(c) . The three possible cases are shown as comparison of K_(c) and Q_(c) in the following figures. Change in Gibbs free energy, i.e., Delta G is the driving force of any reaction. For spontaneous reaction , Delta G =-ve For non-spontaneous reaction , Delta G=+ve For reaction at equilibrium , Delta G =0 Thermodynamically, we know that Delta G= Delta G^(@)+ RT ln Q , where Q is reaction quotient and Delta G^(@)= change in Gibbs energy at standard condition. For equilibrium A(g) hArr B(g) (K_(eq) =1.732) If the pressure of the system [varied by introducing a stream of A (g) and B (g) is represented by the curve at constant temperature T. Suppose N_(2)O_(4)(g) is enclosed in a cylinder fitted with a movable piston which attains the following equilibrium N_(2)O_4)(g) hArr 2NO_(2)(g) Given that for the 10 atmosphere pressure of the equilibrium mixture, the content of NO_(2) is 8xx1^(5) ppm. if the piston of cylinder is moved upward in such a manner so that the volume of the gaseous mixture becomes double, then what will be new ppm of NO_(2)(g) in the cylinder ? ( Assuming that the temperature of the cylinder remains constant )

    A
    `8.2 xx 10^(5) `ppm
    B
    `10^(5)` ppm
    C
    `8.72 xx 10^(5)` ppm
    D
    `7.4 xx 10^(5)` ppm

  • The reaction which is in dynamic equilibrium, ensured us, that the reaction is reversible . But if that the reaction is in equilibrium. The reaction quotient predict either the reversible reaction is in equilibrium or tries to achieve equilibrium. In those reactions which have not achieved equilibrium, we obtain reaction quotient Q_(c) in place of equilibrium constant (K_(c)) by substituting the concentration of reactant and product at the time, at whih we have to calculate the value of Q_(c) . To determine the direction at which the net reaction will proceed to achieve equilibrium, we compare values of Q_(c) and K_(c) . The three possible cases are shown as comparison of K_(c) and Q_(c) in the following figures. Change in Gibbs free energy, i.e., Delta G is the driving force of any reaction. For spontaneous reaction , Delta G =-ve For non-spontaneous reaction , Delta G=+ve For reaction at equilibrium , Delta G =0 Thermodynamically, we know that Delta G= Delta G^(@)+ RT ln Q , where Q is reaction quotient and Delta G^(@)= change in Gibbs energy at standard condition. For equilibrium A(g) hArr B(g) (K_(eq) =1.732) If the pressure of the system [varied by introducing a stream of A (g) and B (g) is represented by the curve at constant temperature T. For the equilibrium n Butane (g) hArr Isobutane (g),(K_(eq)=1.732) If the pressure of the system (varied by introducing a stream of n butane and isobutane ) is represented by the curve at constant temperature T. At a particular point Q, which of the following statements holds good ?

    A
    The reaction moves in the backward direction.
    B
    The reaction movess in the forward direction.
    C
    The reaction is at equilibrium .
    D
    The data is insufficient to prodict

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    The reaction which is in dynamic equilibrium, ensured us, that the reaction is reversible . But if that the reaction is in equilibrium. The reaction quotient predict either the reversible reaction is in equilibrium or tries to achieve equilibrium. In those reactions which have not achieved equilibrium, we obtain reaction quotient Q_(c) in place of equilibrium constant (K_(c)) by substituting the concentration of reactant and product at the time, at whih we have to calculate the value of Q_(c) . To determine the direction at which the net reaction will proceed to achieve equilibrium, we compare values of Q_(c) and K_(c) . The three possible cases are shown as comparison of K_(c) and Q_(c) in the following figures. Change in Gibbs free energy, i.e., Delta G is the driving force of any reaction. For spontaneous reaction , Delta G =-ve For non-spontaneous reaction , Delta G=+ve For reaction at equilibrium , Delta G =0 Thermodynamically, we know that Delta G= Delta G^(@)+ RT ln Q , where Q is reaction quotient and Delta G^(@)= change in Gibbs energy at standard condition. For equilibrium A(g) hArr B(g) (K_(eq) =1.732) If the pressure of the system [varied by introducing a stream of A (g) and B (g) is represented by the curve at constant temperature T. For equilibrium system N_(2)O_(4)(g) hArr 2NO_(2) (g) N_(2)O_(4)(g) is in a cylinder which is fitted with movable piston. Assume that equilibrium partial pressure of N_(2)O_(4) (g) and NO_(2)(g) are 10 and 14 atmospheres respectively. If the piston of the cylinder is pulled out in such way so that volume of the system than what will be the value of equilibrium partial pressure of the NO_(2) gas ?

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