For the reaction N2(s)+O2(s) leftharpoons 2 N O(g), the value of Kc at 800° C is 0.1 . When the equilibrium concentrations of both the reactants is 0.5 mol, what is the value of KP at the same temperature?

Question

For the reaction N2(s)+O2(s) leftharpoons 2 N O(g), the value of Kc at 800° C is 0.1 . When the equilibrium concentrations of both the reactants is 0.5 mol, what is the value of KP at the same temperature? (A) 0.5 (B) 0.1 (C) 0.01 (D) 0.025

Tardigrade Admin · Accepted Answer

N 2(g)+ O 2(g) leftharpoons 2 NO (g) Kc=0 . Lp=Kc(R T)Δ n Δ n= Number of gaseous product - Number of gaseous reactants Δ n=2-2=0 Δ n=0 ∴ Kp=Kc=0.1

Q. For the reaction
$N_{2 (s)} + O_{2 (s)} ⇌ 2 N O_{(g)}$ , the value of $K_{c}$ at $80 0^{\circ} C$ is $0.1.$ When the equilibrium concentrations of both the reactants is $0.5 m o l$ , what is the value of $K_{P}$ at the same temperature?

0.5

0.1

0.01

0.025

$N_{2} (g) + O_{2} (g) ⇌ 2 NO (g)$

$K_{c} = 0. L_{p} = K_{c} (RT)^{Δ n}$

$Δ n =$ Number of gaseous product

- Number of gaseous reactants

$Δ n = 2 - 2 = 0 Δ n = 0$

$∴ K_{p} = K_{c} = 0.1$

Q. For the reaction N2(s)​+O2(s)​⇌2NO(g)​, the value of Kc​ at 800∘C is 0.1. When the equilibrium concentrations of both the reactants is 0.5mol, what is the value of KP​ at the same temperature?

0.5

0.1

0.01

0.025

N2​(g)+O2​(g)⇌2NO(g) Kc​=0.Lp​=Kc​(RT)Δn Δn= Number of gaseous product - Number of gaseous reactants Δn=2−2=0Δn=0 ∴Kp​=Kc​=0.1

Q. For the reaction
$N_{2 (s)} + O_{2 (s)} ⇌ 2 N O_{(g)}$ , the value of $K_{c}$ at $80 0^{\circ} C$ is $0.1.$ When the equilibrium concentrations of both the reactants is $0.5 m o l$ , what is the value of $K_{P}$ at the same temperature?

$N_{2} (g) + O_{2} (g) ⇌ 2 NO (g)$

$K_{c} = 0. L_{p} = K_{c} (RT)^{Δ n}$

$Δ n =$ Number of gaseous product

- Number of gaseous reactants

$Δ n = 2 - 2 = 0 Δ n = 0$

$∴ K_{p} = K_{c} = 0.1$