Consider the reaction: 2 NO(g) + Br2(g) ∆ 2 NOBr(g) Kp = In a reaction mixture at equilibrium, the partial pressure of NO is 108 torr and that of Br2 is 126 torr. What is the partial pressure of NOBr in this mixture?

The question is incomplete, here is the complete question:

Consider the following reaction:

2NO(g) +Br₂(g) ⇔ 2NOBr(g)   Kp= 28.4 at 298 K

In a reaction mixture at equilibrium, the partial pressure of NO is 107 torr and that of Br₂ is 166 torr. What is the partial pressure of NOBr in this mixture?

The equilibrium partial pressure of NOBr in the mixture is 6460.53 torr

Explanation:

We are given:

Equilibrium partial pressure of NO = 108 torr

Equilibrium partial pressure of bromine gas = 126 torr

The given chemical equation follows:

The expression of for above equation follows:

We are given:

Putting values in above expression, we get:

Neglecting the negative value because partial pressure cannot be negative.

Hence, the equilibrium partial pressure of NOBr in the mixture is 6460.53 torr

c-14 is carbon-14, or radiocarbon.   it is a radioactive isotope of carbon with an atomic nucleus containing 6 protons and 8 neutrons.