Home
Class 12
CHEMISTRY
We know, DeltaG^(0)=-RTlnK(c) and DeltaG...

We know, `DeltaG^(0)=-RTlnK_(c) and DeltaG^(0)=-RTlnK_(p)`. State whether the value of `DeltaG^(0)` will be same or not provided the numerical values of both `K_(p) and K_(c)` are different.

Text Solution

Verified by Experts

The values of `DeltaG^(0)` will be different. If the numerical values of `K_(p) and K_(c)` differ, then the value of `DeltaG^(0)` will also be Different because for `K_(p)` and `K_(c)` the corresponding standard states are different in the given two cases. When `DeltaG^(0)=-RTlnK_(c)`. the concentrations of all the reactants and products are taken as 1(M) at standard states. Alternatively, when `DeltaG^(0)=-RTlnK_(p)`, the partial pressures of all the reactants and products at standard states are taken as 1 atm.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • EQUILIBRIUM

    CHHAYA PUBLICATION|Exercise ENTRANCE QUESTIONS BANK|81 Videos

  • EQUILIBRIUM

    CHHAYA PUBLICATION|Exercise SINGLE CORRECT TYPE (MCQ HOTSPOT)|84 Videos

  • EQUILIBRIUM

    CHHAYA PUBLICATION|Exercise SOLVED NCERT EXERCISE|97 Videos

  • ENVIRONMENTAL CHEMISTRY

    CHHAYA PUBLICATION|Exercise PRACTICE SET 14(Answer the following questions)|6 Videos

  • GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

    CHHAYA PUBLICATION|Exercise PRACTICE SET 6|10 Videos

Similar Questions

Explore conceptually related problems

We know, DeltaG^(0)=-RTlnK_(c) and DeltaG^(0)=-RTlnK_(p) . Therefore in case of a reaction occurring in gaseous phase at a given temperature, DeltaG^(0) is the same evenn if the values of K_(p) and K_(c) are different. Is the statement true? Given reasons.

Find the value of k if p(x)=x^(2)+kx+6andp(2)=0 .

Knowledge Check

  • For which of the following reactions are the numerical value of K_p and K_c the same ?

    A
    `H_2(g) + Cl_2(g) iff 2HCl(g)`
    B
    `2SO_2(g) + O_2(g) iff 2SO_3(g)`
    C
    `2NOCl(g) iff 2NO(g) + Cl_2(g)`
    D
    `N_2(g)+3H_2(g) iff 2NH_3(g)`

  • If the E_(cell)^(0) for a given reaction has a negative value, which of the following gives the correct relationships for the values of DeltaG^(0) and K_(eq) ?

    A
    `DeltaG^(0) lt K_(eq) lt 1`
    B
    `DeltaG^(0) gt 0, K_(eq) lt 1`
    C
    `DeltaG^(0) gt 0, K_(eq) gt 1`
    D
    `DeltaG^(0) lt 0 , K_(eq) gt 1`

  • Let DeltaG^@ be the difference in free energy of the reaction when all the reactants and products are in the standard state (1 atmospheric pressure and 298K) and K_c and Kp be the thermodynamic equilibrium constant of the reaction. Both are related to each other at temperature T by the following relation : DeltaG^@ = - 2.303 RT log K_c ? and DeltaG^@ = -2,303 RT log K_p (incase of ideal gas) This equation represents one of the most important results of thermodynamics and relates to the equilibrium constant of a reaction to a thermodynamic property. It is sometimes easier to calculate the free energy ina reaction rather than to measure the equilibrium constant. Standard free energy change can be thermodynamically calculated as DeltaG^@ = DeltaH^@- TDeltaS^@ Here DeltaH^@ = standard enthalpy change DeltaS^@ = standard entropy change. At 490^@C , the value of equilibrium constant, K_p is 45.9 the reaction H_2(g) + I_2(g) iff 2Hi(g) Calculate the value of DeltaG^@ for the reaction at that temperature

    A
    -3.5 kcal
    B
    3.5 kcal
    C
    5.79 kcal
    D
    -5.79 kcal

    • Similar Questions

    Explore conceptually related problems

    Relate K_(p) and K_(c) when Deltan=0, Deltan=1, Deltan=2.0

    The value of DeltaG^@ for the phosphorylation of glucose in glycolysis is 13.8 ku/mol. Find the value of K_c at 298 K.

    At constant temperature, if the pressure is changed at equilibrium of a gaseous reaction, then will the values of K_(p),K_(c) and K_(x) change?

    At 298K, for attainment of equilibrium of the reaction N_(2)O_(4)(g)hArr 2NO_(2)(g) , 5 mol of each of the constituents is taken. Due to this, total pressure of the mixture turns 20 atm. If DeltaG_(f)^(0)(N_(2)O_(4))=100kJ*mol^(-1) and DeltaG_(f)^(0)(NO_(2))=50K(J*mol^(-1)) then- (1) Give the value of DeltaG of the reaction? (2) In which direction will the reaction proceed more to attain equilibrium?

    Let DeltaG^@ be the difference in free energy of the reaction when all the reactants and products are in the standard state (1 atmospheric pressure and 298K) and K_c and Kp be the thermodynamic equilibrium constant of the reaction. Both are related to each other at temperature T by the following relation : DeltaG^@ = - 2.303 RT log K_c ? and DeltaG^@ = -2,303 RT log K_p (incase of ideal gas) This equation represents one of the most important results of thermodynamics and relates to the equilibrium constant of a reaction to a thermodynamic property. It is sometimes easier to calculate the free energy ina reaction rather than to measure the equilibrium constant. Standard free energy change can be thermodynamically calculated as DeltaG^@ = DeltaH^@- TDeltaS^@ Here DeltaH^@ = standard enthalpy change DeltaS^@ = standard entropy change. Calculate the equilibrium concentration ratio of C to A lf 2.0 mol each of A and B were allowed to come to equilibrium at 300 K A+B iff , DeltaG^@ = 460 cal

    Home
    Profile