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In an equilibrium reaction for which Del...

In an equilibrium reaction for which `Delta G^(@) = 0` , the equilibrium constant K =

A

0

B

1

C

2

D

10

Text Solution

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The correct Answer is:
To determine the value of the equilibrium constant \( K \) when \( \Delta G^\circ = 0 \), we can follow these steps: ### Step-by-Step Solution: 1. **Understand the relationship between Gibbs free energy and the equilibrium constant:** The Gibbs free energy change (\( \Delta G \)) for a reaction is related to the equilibrium constant (\( K \)) by the equation: \[ \Delta G^\circ = -RT \ln K \] where: - \( R \) is the universal gas constant (8.314 J/(mol·K)), - \( T \) is the temperature in Kelvin, - \( K \) is the equilibrium constant. 2. **Set \( \Delta G^\circ \) to zero:** Given that \( \Delta G^\circ = 0 \), we can substitute this value into the equation: \[ 0 = -RT \ln K \] 3. **Rearrange the equation:** Since \( -RT \) is not zero (as long as \( R \) and \( T \) are positive), we can divide both sides by \( -RT \): \[ \ln K = 0 \] 4. **Solve for \( K \):** To find \( K \), we need to exponentiate both sides: \[ K = e^0 \] Since \( e^0 = 1 \): \[ K = 1 \] ### Final Answer: Thus, when \( \Delta G^\circ = 0 \), the equilibrium constant \( K \) is equal to 1. ---
To determine the value of the equilibrium constant \( K \) when \( \Delta G^\circ = 0 \), we can follow these steps: ### Step-by-Step Solution: 1. **Understand the relationship between Gibbs free energy and the equilibrium constant:** The Gibbs free energy change (\( \Delta G \)) for a reaction is related to the equilibrium constant (\( K \)) by the equation: \[ \Delta G^\circ = -RT \ln K ...
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Knowledge Check

  • In an equilibrium reaction for which DeltaG^(ɵ)=0 , the equilibrium constant K should be equal to :

    A
    Zero
    B
    `10`
    C
    `1`
    D
    `2`

  • In an equilibrium reaction for which DeltaG=0, the equlibrium constant K =

    A
    0
    B
    1
    C
    2
    D
    10

  • If the equilibrium constant for a reaction is 4*0 , what will be the equilibrium constant for the reverse reaction.

    A
    1
    B
    4
    C
    0.25
    D
    25

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    For the reaction H_(2)(g)+I_(2)(g)hArr 2HI(g) , the standard free energy is Delta G^(Θ) gt 0 . The equilibrium constant (K) would be........ .

    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 ?

    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 ?

    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 ?

    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 )

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