Equilibrium constant KC for the following reaction at 800 K is, 4 NH3(g)leftharpoons(1/2)N2(g)+(3/2)H2(g) The value of KP for the following reaction will be N2(g)+3H2(g)leftharpoons2NH3(g)

Question

Equilibrium constant KC for the following reaction at 800 K is, 4 NH3(g)leftharpoons(1/2)N2(g)+(3/2)H2(g) The value of KP for the following reaction will be N2(g)+3H2(g)leftharpoons2NH3(g) (A) ((800 R/4))2 - (B) 16× (800R)2 (C) [(1/4 × 800 R)]2 (D) (800 R)1 / 24

Tardigrade Admin · Accepted Answer

NH 3( g ) leftharpoons (1/2) N 2( g )+(3/2) H 2( g ) ; K C =4 N 2( g )+3 H 2( g ) leftharpoons 2 NH 3( g ) Δ n g =2-4=-2 So KC for the following reaction will be equal to (1/42) K p = K C ( RT )Δ n g =(1/42) ×(800 × R )-2 =(1/4 × 800 R ) 2

Q. Equilibrium constant $K_{C}$ for the following reaction at $800 K$ is, $4$
$N H_{3} (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g)$
The value of $K_{P}$ for the following reaction will be
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$

$(\frac{800 R}{4})^{2 -}$

$16 \times (800 R)^{2}$

$[\frac{1}{4 \times 800 R}]^{2}$

$(800 R)^{1/2} 4$

Q. Equilibrium constant KC​ for the following reaction at 800K is, 4 NH3​(g)⇌21​N2​(g)+23​H2​(g) The value of KP​ for the following reaction will be N2​(g)+3H2​(g)⇌2NH3​(g)

(4800R​)2−

16×(800R)2

[4×800R1​]2

(800R)1/24

NH3​(g)⇌21​N2​(g)+23​H2​(g);KC​=4 N2​(g)+3H2​(g)⇌2NH3​(g)Δng​=2−4=−2 So KC​ for the following reaction will be equal to 421​ Kp​=KC​(RT)Δng​ =421​×(800×R)−2 =4×800R1​2

Q. Equilibrium constant $K_{C}$ for the following reaction at $800 K$ is, $4$
$N H_{3} (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g)$
The value of $K_{P}$ for the following reaction will be
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$

$(\frac{800 R}{4})^{2 -}$

$16 \times (800 R)^{2}$

$[\frac{1}{4 \times 800 R}]^{2}$

$(800 R)^{1/2} 4$