Question

At 497 K, this reaction has a Kc value of 0.03902X(g)+2Y(g) <----> 2Z(g)Calculate Kp at 497 K

Answer 1

Answer:

2.34 × 10⁻⁵

Explanation:

Let's consider the following reaction at equilibrium.

2 X(g) + 2 Y(g) ⇄ 2 Z(g)

The concentration equilibrium constant (Kc) has a value of 0.0390. We can find the pressure equilibrium constant (Kp) using the following expression.

$Kp=Kc(RT)^{\Delta n(g) }$

where,

R is the ideal gas constant (0.08206 atm.L/mol.K)

T is the absolute temperature

Δn(g) = (Total moles of gas on the products side) - (Total moles of gas on the reactants side) = 2 - 4 = -2

$Kp=0.0390(0.08206 \times 497)^{-2}=2.34 \times 10^{-5}$