At 1000 K, the equilibrium constant. KC for the reaction 2 NOCl (g) leftharpoons 2 NO (g)+ Cl 2(g) is 4.0 × 10-6 mol L -1. The KP (in bar) at the same temperature is (R=0.083 L. bar . K -1 mol -1)

Question

At 1000 K, the equilibrium constant. KC for the reaction 2 NOCl (g) leftharpoons 2 NO (g)+ Cl 2(g) is 4.0 × 10-6 mol L -1. The KP (in bar) at the same temperature is (R=0.083 L. bar . K -1 mol -1) (A) 3.32 × 10-6 (B) 3.32 × 104 (C) 3.32 × 10-4 (D) 3.32 × 10-3

Tardigrade Admin · Accepted Answer

Given, KC=4 × 10-6 mol / L 2 NOCl (g) leftharpoons 2 NO (g)+ Cl 2(g) Δ n= product mole - reactant mole =3-2=1 From Kp equation Kp=KC ×(R T)Δ n =4 × 10-6 mol / L ×(0.0831 L text bar k -1 mol -1. × 1000 K )2 Kp=3.32 × 10-4 bar

Q. At $1000 K$ , the equilibrium constant. $K_{C}$ for the reaction $2 NOCl (g) ⇌ 2 NO (g) + C l_{2} (g)$ is $4.0 \times 1 0^{- 6} m o l L^{- 1}$ . The $K_{P}$ (in bar) at the same temperature is $(R = 0.083 L$ bar $K^{- 1} m o l^{- 1})$

$3.32 \times 1 0^{- 6}$

$3.32 \times 1 0^{4}$

$3.32 \times 1 0^{- 4}$

$3.32 \times 1 0^{- 3}$

Q. At 1000K, the equilibrium constant. KC​ for the reaction 2NOCl(g)⇌2NO(g)+Cl2​(g) is 4.0×10−6molL−1. The KP​ (in bar) at the same temperature is (R=0.083L bar K−1mol−1)

3.32×10−6

3.32×104

3.32×10−4

3.32×10−3

Given, KC​=4×10−6mol/L 2NOCl(g)⇌2NO(g)+Cl2​(g) Δn= product mole − reactant mole =3−2=1 From Kp​ equation Kp​=KC​×(RT)Δn =4×10−6mol/L ×(0.0831L bar k−1mol−1×1000K)2 Kp​=3.32×10−4 bar

Q. At $1000 K$ , the equilibrium constant. $K_{C}$ for the reaction $2 NOCl (g) ⇌ 2 NO (g) + C l_{2} (g)$ is $4.0 \times 1 0^{- 6} m o l L^{- 1}$ . The $K_{P}$ (in bar) at the same temperature is $(R = 0.083 L$ bar $K^{- 1} m o l^{- 1})$

$3.32 \times 1 0^{- 6}$

$3.32 \times 1 0^{4}$

$3.32 \times 1 0^{- 4}$

$3.32 \times 1 0^{- 3}$