Calculate the equilibrium constant kp for this reaction, given the following information (at 299 k ): 2no(g)+br2(g)⇌2nobr(g)kc=2.2 2no(g)⇌n2(g)+o2(g)kc=2.3×1030

, Kp = 0.08967

, Kp = 2.3×10³⁰

Explanation:

The relation between Kp and Kc is given below:

Where,

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

Given: Kc = 2.2

Temperature = 299 K

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (2)-(2+1) = -1

Thus, Kp is:

Kp = 0.08967

For the second equilibrium reaction:

Given: Kc = 2.3×10³⁰

Temperature = 299 K

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (2)-(2) = 0

Thus, Kp is:

Kp =  2.3×10³⁰

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.