At 350 K, Kp for the reaction given below is 3.0 × 1010 bar-1 at equilibrium. What will be the value of Kc at this temperature? 2N2(g) +O2(g) leftharpoons 2N2O(g)

Question

At 350 K, Kp for the reaction given below is 3.0 × 1010 bar-1 at equilibrium. What will be the value of Kc at this temperature? 2N2(g) +O2(g) leftharpoons 2N2O(g) (A) 7.4×1011 L mol-1 (B) 8715×1010 L mol-1 (C) 0.08 L mol-1 (D) 8715×1011 L mol-1

Tardigrade Admin · Accepted Answer

Kp=Kc(RT)Δ n ; Δ n=2-3=-1 T=350 K, R=0.083 bar L K-1 mol-1 Kc=(Kp/(RT)Δ n) Kc=(3×1010 bar-1/(0.083 L bar K-1 mol-1×350 K)-1) =3×1010 bar-1×0.083 L bar K-1 mol-1×350 K =8715×1011 L mol-1

Q. At $350 K$ , $K_{p}$ for the reaction given below is $3.0 \times 1 0^{10} ba r^{- 1}$ at equilibrium. What will be the value of $K_{c}$ at this temperature?
$2 N_{2 (g)} + O_{2 (g)} ⇌ 2 N_{2} O_{(g)}$

$7.4 \times 1 0^{11} L m o l^{- 1}$

$8715 \times 1 0^{10} L m o l^{- 1}$

$0.08 L m o l^{- 1}$

$8715 \times 1 0^{11} L m o l^{- 1}$

Q. At 350K, Kp​ for the reaction given below is 3.0×1010bar−1 at equilibrium. What will be the value of Kc​ at this temperature? 2N2(g)​+O2(g)​⇌2N2​O(g)​

7.4×1011Lmol−1

8715×1010Lmol−1

0.08Lmol−1

8715×1011Lmol−1

Kp​=Kc​(RT)Δn ; Δn=2−3=−1 T=350K, R=0.083barLK−1mol−1 Kc​=(RT)ΔnKp​​ Kc​=(0.083LbarK−1mol−1×350K)−13×1010bar−1​ =3×1010bar−1×0.083LbarK−1mol−1×350K =8715×1011Lmol−1

Q. At $350 K$ , $K_{p}$ for the reaction given below is $3.0 \times 1 0^{10} ba r^{- 1}$ at equilibrium. What will be the value of $K_{c}$ at this temperature?
$2 N_{2 (g)} + O_{2 (g)} ⇌ 2 N_{2} O_{(g)}$

$7.4 \times 1 0^{11} L m o l^{- 1}$

$8715 \times 1 0^{10} L m o l^{- 1}$

$0.08 L m o l^{- 1}$

$8715 \times 1 0^{11} L m o l^{- 1}$