At 600 K 200 bar, a 1: 3 (molar ratio) mixture of A 2 B 2 react to form an equilibrium mixture containing chi AB 3=0.60 Assuming ideal gas behaviour, calculate K p for the reaction: A 2( g )+3 B 2( g ) leftharpoons 2 AB 3( g )

Question

At 600 K 200 bar, a 1: 3 (molar ratio) mixture of A 2 B 2 react to form an equilibrium mixture containing chi AB 3=0.60 Assuming ideal gas behaviour, calculate K p for the reaction: A 2( g )+3 B 2( g ) leftharpoons 2 AB 3( g ) (A) 3.33 × 10-3 (B) 3.33 × 10-11 (C) 5.61 × 10-7 (D) 9.87 × 10-12

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/9900b0eae827e3196ec827cf0c0899db-.png" /> (∵ total no. of moles = n (1- x )+3 n (1- x )+2 nx = n (4-2 x)) Also, chi AB 3=(2 x /4-2 x )=0.6 x=0.75 So, chiA2=0.1 XB2=0.3 Use the expression, Kp=((p A B3)2/(p A2)(p B2)3) =3.33 × 10-3

Q. At $600 K$ & $200$ bar, a $1 : 3$ (molar ratio) mixture of $A_{2}$ & $B_{2}$ react to form an equilibrium mixture containing $χ_{A B_{3}} = 0.60$ Assuming ideal gas behaviour, calculate $K_{p}$ for the reaction:
$A_{2} (g) + 3 B_{2} (g) ⇌ 2 A B_{3} (g)$

$3.33 \times 1 0^{- 3}$

$3.33 \times 1 0^{- 11}$

$5.61 \times 1 0^{- 7}$

$9.87 \times 1 0^{- 12}$

$(∵$ total no. of moles $= n (1 - x) + 3 n (1 - x) + 2 n x = n (4 - 2 x))$

Also, $χ_{A B_{3}} = \frac{2 x}{4 - 2 x} = 0.6$

$x = 0.75$

So, $χ_{A_{2}} = 0.1& X_{B_{2}} = 0.3$

Use the expression, $K_{p} = \frac{( p A B _{3} ) ^{2}}{( p A _{2} ) ( p B _{2} ) ^{3}}$

$= 3.33 \times 1 0^{- 3}$

Q. At 600K & 200 bar, a 1:3 (molar ratio) mixture of A2​ & B2​ react to form an equilibrium mixture containing χAB3​​=0.60 Assuming ideal gas behaviour, calculate Kp​ for the reaction: A2​(g)+3B2​(g)⇌2AB3​(g)

3.33×10−3

3.33×10−11

5.61×10−7

9.87×10−12