Question

For the reaction \[\ce{A(g) <=> 2B(g)}\] 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]

K_p for the reaction at 1000 K is:

Options

• 83.1

• 2.077 × 10⁵

• 0.033

• 0.021

Answer 1

0.033

Explanation:

K_c = `K_f/K_b`

= `K_f/(2500 K_f)`

= `1/2500`

K_p = K_c(RT)^Δng

= `1/2500 (0.0831 xx 1000)`

= 0.033