For reaction, 2 NOCl ( g ) leftharpoons 2 NO (g)+ Cl 2(g), KC at 427° C is 3 × 10-6 L ⋅ mol -1 . The value of Kp is nearly

Question

For reaction, 2 NOCl ( g ) leftharpoons 2 NO (g)+ Cl 2(g), KC at 427° C is 3 × 10-6 L ⋅ mol -1 . The value of Kp is nearly (A) 750 × 10-5 (B) 2.50 × 10-5 (C) 2.50 × 10-4 (D) 1.75 × 10-4 .

Tardigrade Admin · Accepted Answer

Kp=Kc(R T)Δ n Δ n= change in gaseous moles. For the reaction 2 NOCl (g) leftharpoons2 NO (g)+ Cl 2(g) Kc=3 × 10-6 L mol -1, R=0.0821 L atm / K mol Δ n=3-2=1 T=427+273=700 K Kp=3 × 10-6 ×(0.0821 × 700)1 =1.72 × 10-4

Q. For reaction, $2 NOC l_{(g)} ⇌ 2 N O_{(g)} + C l_{2} (g), K_{C}$ at $42 7^{\circ} C$ is $3 \times 1 0^{- 6} L \cdot m o l^{- 1} .$ The value of $K_{p}$ is nearly

$750 \times 1 0^{- 5}$

$2.50 \times 1 0^{- 5}$

$2.50 \times 1 0^{- 4}$

$1.75 \times 1 0^{- 4} .$

Q. For reaction, 2NOCl(g)​⇌2NO(g)​+Cl2​(g),KC​ at 427∘C is 3×10−6L⋅mol−1. The value of Kp​ is nearly

750×10−5

2.50×10−5

2.50×10−4

1.75×10−4.

Kp​=Kc​(RT)Δn Δn= change in gaseous moles. For the reaction 2NOCl(g)⇌2NO(g)+Cl2​(g) Kc​=3×10−6Lmol−1,R=0.0821Latm/Kmol Δn=3−2=1 T=427+273=700K Kp​=3×10−6×(0.0821×700)1 =1.72×10−4

Q. For reaction, $2 NOC l_{(g)} ⇌ 2 N O_{(g)} + C l_{2} (g), K_{C}$ at $42 7^{\circ} C$ is $3 \times 1 0^{- 6} L \cdot m o l^{- 1} .$ The value of $K_{p}$ is nearly

$750 \times 1 0^{- 5}$

$2.50 \times 1 0^{- 5}$

$2.50 \times 1 0^{- 4}$

$1.75 \times 1 0^{- 4} .$