Question

Consider the reaction: I2(g)+Cl2(g)⇌2ICl(g) Kp= 81.9 at 25 ∘C. Calculate ΔGrxn for the reaction at 25 ∘C under each condition: -(Part 1) At standard conditions -(Part 2) PICl= 2.63atm ; PI2= 0.324atm ; PCl2= 0.217atm .

Answer 1

Answer:

Part 1: - 1.091 x 10⁴ J/mol.

Part 2: - 1.137 x 10⁴ J/mol.

Explanation:

Part 1: At standard conditions:

At standard conditions Kp= 81.9.

∵ ΔGrxn = -RTlnKp

∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(81.9)) = - 1.091 x 10⁴ J/mol.

Part 2: PICl = 2.63 atm; PI₂ = 0.324 atm; PCl₂ = 0.217 atm.

For the reaction:

I₂(g) + Cl₂(g) ⇌ 2ICl(g).

Kp = (PICl)²/(PI₂)(PCl₂) = (2.63 atm)²/(0.324 atm)(0.217 atm) = 98.38.

∵ ΔGrxn = -RTlnKp

∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(98.38)) = - 1.137 x 10⁴ J/mol.

Answer 2

a. ΔG°rxn at standard conditions : -1.09.10⁴ J/mol

b. ΔG°rxn at  PICl= 2.63atm ; PI2= 0.324atm ; PCl2= 0.217atm  =  -1.14.10⁴ J/mol

Further explanation

The equilibrium constant is the ratio of the concentration or pressure between the result of the reaction / product and the reactant with each reaction coefficient raised

The equilibrium constant is based on the concentration (Kc) in a reaction

pA + qB -----> mC + nD

$\large {\boxed {\bold {Kc ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}$

Whereas the equilibrium constant is based on partial pressure  (Kp)

$\large {\boxed {\bold {Kp ~ = ~ \frac {[pC] ^ m [pD] ^ n} {[pA] ^ p [pB] ^ q}}}}$

The value of Kp and Kc can be linked to the formula

$\large {\boxed {\bold {Kp ~ = ~ Kc. (R.T) ^ {\Delta n}}}}$

R = gas constant = 0.0821 L.atm / mol.K

Δn = number of product coefficients-number of reactant coefficients

Gibbs (Free) Energy = ΔGrxn when connected to Kc / Kp:

ΔG°rxn is the standard Gibbs free energy of reaction,

ΔGrxn = ΔG°rxn + RTlnQ

At equilibrium, Q = K and ΔGrxn = 0

ΔG°rxn = -RTlnK

• a. Standard condition  (T = 298 K)

ΔG°rxn = -RTlnK

ΔG°rxn = - 8,314. 298 ln 81.9 = -10914.94 J / mol  = -1.09.10⁴ J/mol

• b. PICl = 2.63 atm; PI2 = 0.324 atm; PCl2 = 0.217 at m.

$Kp=\frac{PICl^2}{(PI_2)(PCl_2)}$

$= \frac{2.63^2}{0.324.0.217}$

Kp = 98.3799

ΔG°rxn = -RTlnK

ΔG°rxn =  - 8,314. 298 ln 98.3799

ΔG°rxn = -11369.1728 J/mol = -1.14.10⁴ J/mol

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Keywords : kc, kp, ΔGrxn,Gibbs (Free) Energy, standard conditions