The standard free energy change (Δ G °) for 50 % dissociation of N 2 O 4 into NO 2 at 27° C and 1 atm pressure is -x J mol -1. The value of x is (Nearest Integer) [Given: R =8.31 J K -1 mol -1, log 1.33=0.1239 ln 10=2.3]

Question

The standard free energy change (Δ G °) for 50 % dissociation of N 2 O 4 into NO 2 at 27° C and 1 atm pressure is -x J mol -1. The value of x is (Nearest Integer) [Given: R =8.31 J K -1 mol -1, log 1.33=0.1239 ln 10=2.3] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/55cda320d20a5fe3ca1262b991879877-.png" /> k P =(((1/1.5) × 1)2/((0.5)1.5 × 1))=(1/0.75)=(100/75) =1.33 Δ G 0=- RT ℓ nk P =-8.31 × 300 × ℓ n (1.33)=-710.45 J / mol =-710 J / mol .

Q. The standard free energy change $(Δ G^{\circ})$ for $50%$ dissociation of $N_{2} O_{4}$ into $N O_{2}$ at $2 7^{\circ} C$ and $1$ atm pressure is $- x J m o l^{- 1}$ . The value of $x$ is ________(Nearest Integer)
[Given : $R = 8.31 J K^{- 1} m o l^{- 1}, lo g 1.33 = 0.1239$ $ln 10 = 2.3]$

$k_{P} = \frac{( \frac{1}{1.5} \times 1 ) ^{2}}{( \frac{0.5}{1.5} \times 1 )} = \frac{1}{0.75} = \frac{100}{75}$
$= 1.33$
$Δ G^{0} = - RT ℓ n k_{P}$
$= - 8.31 \times 300 \times ℓ n (1.33) = - 710.45 J / m o l$
$= - 710 J / m o l .$

Q. The standard free energy change (ΔG∘) for 50% dissociation of N2​O4​ into NO2​ at 27∘C and 1 atm pressure is −xJmol−1. The value of x is ________(Nearest Integer) [Given : R=8.31JK−1mol−1,log1.33=0.1239 ln10=2.3]

kP​=(1.50.5​×1)(1.51​×1)2​=0.751​=75100​ =1.33 ΔG0=−RTℓnkP​ =−8.31×300×ℓn(1.33)=−710.45J/mol =−710J/mol.

Q. The standard free energy change $(Δ G^{\circ})$ for $50%$ dissociation of $N_{2} O_{4}$ into $N O_{2}$ at $2 7^{\circ} C$ and $1$ atm pressure is $- x J m o l^{- 1}$ . The value of $x$ is ________(Nearest Integer)
[Given : $R = 8.31 J K^{- 1} m o l^{- 1}, lo g 1.33 = 0.1239$ $ln 10 = 2.3]$

$k_{P} = \frac{( \frac{1}{1.5} \times 1 ) ^{2}}{( \frac{0.5}{1.5} \times 1 )} = \frac{1}{0.75} = \frac{100}{75}$
$= 1.33$
$Δ G^{0} = - RT ℓ n k_{P}$
$= - 8.31 \times 300 \times ℓ n (1.33) = - 710.45 J / m o l$
$= - 710 J / m o l .$