At 527° C, the reaction given below has K c=4 NH 3( g ) leftharpoons (1/2) N 2( g )+(3/2) H 2( g ) What is the K p for the reaction? N 2( g )+3 H 2( g ) leftharpoons 2 NH 3( g )

Question

At 527° C, the reaction given below has K c=4 NH 3( g ) leftharpoons (1/2) N 2( g )+(3/2) H 2( g ) What is the K p for the reaction? N 2( g )+3 H 2( g ) leftharpoons 2 NH 3( g ) (A) 16 ×(800 R )2 (B) ((800 R /4))-2 (C) ((1/4 × 800 R ))2 (D) None of these

Tardigrade Admin · Accepted Answer

If, Kc=([N2]1 / 2[H2]3 / 2/[N H3]1)=4 then ([ NH 3]2/[ N 2][ H 2]3)=(4)-2=(1/16) also Kp=Kc ⋅(R T)Δ ng

Q. At $52 7^{\circ} C$ , the reaction given below has $K_{c} = 4$
$N H_{3} (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g)$
What is the $K_{p}$ for the reaction?
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$

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

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

$(\frac{1}{4 \times 800 R})^{2}$

None of these

If, $K_{c} = \frac{[ N _{2} ] ^{1/2} [ H _{2} ] ^{3/2}}{[ N H _{3} ] ^{1}} = 4$

then $\frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}} = (4)^{- 2} = \frac{1}{16}$

also $K_{p} = K_{c} \cdot (RT)^{Δ n_{g}}$

Q. At 527∘C, the reaction given below has Kc​=4 NH3​(g)⇌21​N2​(g)+23​H2​(g) What is the Kp​ for the reaction? N2​(g)+3H2​(g)⇌2NH3​(g)

16×(800R)2

(4800R​)−2

(4×800R1​)2