For the given equilibrium reaction, 2 A ( g ) leftharpoons 2 B ( g )+ C ( g ) the equilibrium constant (Kc) at 1000 K is 4 × 10-4. Calculate Kp for the reaction at 800 K temperature.

Question

For the given equilibrium reaction, 2 A ( g ) leftharpoons 2 B ( g )+ C ( g ) the equilibrium constant (Kc) at 1000 K is 4 × 10-4. Calculate Kp for the reaction at 800 K temperature. (A) 0.044 (B) 0.026 (C) 0.33 (D) 1

Tardigrade Admin · Accepted Answer

2 A(g) leftharpoons 2 B(g)+C(g) Kp=KC ⋅(R T)Δ ng Given, KC=4 × 10-4 T=1000 K Δ ng= moles of product - moles of reactant =3-2=1 Kp=4 × 10-4(8.314 × 800)1 Kp=0.026

Q. For the given equilibrium reaction, $2 A (g) ⇌ 2 B (g) + C (g)$ the equilibrium constant $(K_{c})$ at $1000 K$ is $4 \times 1 0^{- 4}$ . Calculate $K_{p}$ for the reaction at $800 K$ temperature.

0.044

0.026

0.33

1

$2 A (g) ⇌ 2 B (g) + C (g)$
$K_{p} = K_{C} \cdot (RT)^{Δ n_{g}}$
Given, $K_{C} = 4 \times 1 0^{- 4}$
$T = 1000 K$
$Δ n_{g} =$ moles of product $-$ moles of reactant
$= 3 - 2 = 1$
$K_{p} = 4 \times 1 0^{- 4} (8.314 \times 800)^{1}$
$K_{p} = 0.026$

Q. For the given equilibrium reaction, 2A(g)⇌2B(g)+C(g) the equilibrium constant (Kc​) at 1000K is 4×10−4. Calculate Kp​ for the reaction at 800K temperature.

0.044

0.026

0.33

1

2A(g)⇌2B(g)+C(g) Kp​=KC​⋅(RT)Δng​ Given, KC​=4×10−4 T=1000K Δng​= moles of product − moles of reactant =3−2=1 Kp​=4×10−4(8.314×800)1 Kp​=0.026

Q. For the given equilibrium reaction, $2 A (g) ⇌ 2 B (g) + C (g)$ the equilibrium constant $(K_{c})$ at $1000 K$ is $4 \times 1 0^{- 4}$ . Calculate $K_{p}$ for the reaction at $800 K$ temperature.

$2 A (g) ⇌ 2 B (g) + C (g)$
$K_{p} = K_{C} \cdot (RT)^{Δ n_{g}}$
Given, $K_{C} = 4 \times 1 0^{- 4}$
$T = 1000 K$
$Δ n_{g} =$ moles of product $-$ moles of reactant
$= 3 - 2 = 1$
$K_{p} = 4 \times 1 0^{- 4} (8.314 \times 800)^{1}$
$K_{p} = 0.026$