Question

.
For the homogenous gaseous reaction A+B=2C+D at 500 K, K = 0.04 bar. The K of the
reaction at 500 K is approximately :
1) 1 x 10-3
2) 4 x 10-2
3) 1 x 10-5
4) 1.6 × 102 mol L-1​

Answer 1

Answer:

For the homogenous gaseous reaction A+B=2C+D at 500 K, K = 0.04 bar. The K of the. reaction at 500 K is approximately : 1) 1 x 10-3.

Answer 2

The equilibrium constant at 500K is 1 × 10⁻³ hence option (1) is correct.

For the homogenous gaseous reaction, A + B ⇒ 2C + D at 500K, Kp = 0.04 bar.

We have to find the equilibrium constant, Kc.

Here chemical reaction, A + B ⇒ 2C + D

Δn = no of moles of products - no of moles of reactants

     = (2 + 1) - (1 + 1) = 1

For chemical reaction,

$K_p=K_c(R T)^{\Delta n}$

$K_p=0.04bar, R=0.082Latm/mol/K, T=500K,\Delta n=1$

⇒0.04 = Kc(0.082 × 500)¹

⇒ kc = 0.04/(0.082 × 500) = 0.000975 ≈ 1 × 10⁻³.

Therefore the equilibrium constant at 500K is 1 × 10⁻³.

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