For the reaction A(g) ⇌ 2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K.
[Given: R = 0.0831 L atm mol⁻¹ K⁻¹]
Kₚ for the reaction at 1000 K is

CONCEPT:

Relation Between K_C and K_P

• The equilibrium constant Kp is related to Kc by the equation:

  K_p = K_C (RT)^Δn_g

• Where:
  • K_C is the equilibrium constant in terms of concentration.
  • R is the gas constant (0.0831 L atm mol^-1 K^-1).
  • T is the temperature in Kelvin (1000 K in this case).
  • Δn_g is the change in the number of moles of gas (products - reactants).
• For the given reaction, the backward rate constant is 2500 times the forward rate constant, and we are asked to calculate Kp at 1000 K.

EXPLANATION:

• The relationship between the rate constants for the forward and backward reactions is:

  \(KC = \frac{k{\text{f}}}{k{\text{b}}}\)

  Given that the backward rate constant is 2500 times the forward rate constant, we have:

  \(KC = \frac{k{\text{f}}}{2500 k{\text{f}}} = \frac{1}{2500}\)

• Next, we use the equation to find Kp :

  K_p = K_C (RT)^Δn_g

  Given that Δn_g = 1 (because there is a change in the number of moles from 1 mole of reactants to 2 moles of products), we substitute the known values:

  \(Kp = \frac{1}{2500} (0.0831 × 1000)^1\)

• Solving the equation:

  \(Kp = \frac{1}{2500} \times 83.1 = 0.033\)

Therefore, the correct value of K_p is 0.033.