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Home/ Questions/Consider the following reaction: I2(g)+Cl2(g) ⇌ 2ICl(g) Kp= 81.9 at 25 ∘ C.
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Asked: April 11, 20222022-04-11T19:13:51+00:00 2022-04-11T19:13:51+00:00In: Chemistry

Consider the following reaction: I2(g)+Cl2(g) ⇌ 2ICl(g) Kp= 81.9 at 25 ∘ C.

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Consider the following reaction:

I2(g) + Cl2(g) ⇌  2ICl(g) Kp= 81.9 at 25 ∘ C.

Calculate ΔGorxn for the reaction at 25 oC under each of the following conditions.

Part A: Standard conditions

Part B: At equilibrium

Part C:

PICl= 2.59 atm

PI2= 0.322 atm

PCl2= 0.221 atm

♦ Relevant knowledge
Free Energy

The Gibbs free energies and the equilibrium constant in relation to the pressure or concentration for the particular species when at particular temperature are correlated to one another according to the equation below. Any undetermined value can be discovered in the event that the other equation parameters are established.

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    Jake Jones
    2022-04-12T00:14:16+00:00Added an answer on April 12, 2022 at 12:14 am
    Concepts and Reason

    This is the reaction:

    I2(g) + Cl2(g) ⇌ 2ICl(g)

    Use the following formula to calculate the Gibbs-free energy change for the reaction.

    ΔGrxn = ΔGo + RT lnQ

    Here, ΔGo is Gibbs free energy change in standard conditions, R gas constant, T is temperature, and Q is reaction quotient.

    Calculate ΔGo by using the following formula.

    ΔGo = -RT lnK

    Here K is the equilibrium constant for the reaction.

    Below, calculate the reaction quotient.

    Ра
Q
P, Ра,
2
ICI
2

    Fundamentals

    Standard free energy change is the change in Gibbs free energy when one mole is made from its pure elements under normal conditions. Standard conditions are 1atm temperature and 298K temperatures.

    At equilibrium, Gibbs free energy for a reaction will equal zero.

    A)

    ΔGo = -RT lnKp

    ΔGo = -8.314 J/mol.K x 298K x ln(81.9)

             = -8.314 J/mol.K x 298 x 4.405

             = -10913 J/mol

             = -10.91 kJ/mol

    -10.91 kJ/mol is therefore standard Gibbs energy change free.

    B)

    At equilibriumGibbs free energy change is zero.

    ΔGo  = 0

    (C)

    2
ICI
Pa
(2.59 atm)
(0.221 atm)x(0.322 atm)
=94.26

    Now calculate AG by substituting -10.91 kJ/mol for AG°, 8.314 J/mol.K for R,
298K for T, and 94.26 for Q.
AG
гхn
-24.8 kJ/mol Answer: 

    Part A:

    Standard Gibbs energy change is -10.91 kJ/mol.

    Part B:

    ΔGo  = 0

    Part C:

    ΔGrxn  = 0.353 kJ/mol

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