0.954 mol bromine is added to 1 L reaction vessel that contains 1.22 mole of hydrogen and 1 mole of HBr at 1500K. What is the partial pressure of hydrogen, bromine and HBr at equilibrium?

Partial pressure of H₂ →  149.9 atm

Partial pressure of Br₂ → 117.2 atm

Partial pressure of HBr → 112.8 atm

Explanation:

The problem can be solved by the Ideal Gases Law.

In a mixture of gases we know that the sum of partial pressures values the total pressure in the vessel. And the total pressure in the vessel is generated by the total moles which are presennt.

Total moles = 0.954 moles + 1.22 moles + 1 mol = 3.174 moles

Let's apply the Law to determine the total pressure.

P . 1 L = 3.174 moles . 0.082 . 1500 K

P = 390.4 atm

Now we apply the mole fraction concept

Mole fraction (1) →  moles of gas / total moles

Mole fraction (2) → partial pressure of gas / total pressure

In conclussion 1 = 2

Partial pressure of H₂ →  (1.22 / 3.174) . 390.4 atm = 149.9 atm

Partial pressure of Br₂ → (0.954 / 3.174) . 390.4 atm = 117.2 atm

Partial pressure of HBr → (1 / 3.174) . 390.4 atm = 112.8 atm

elements make up everything, and they combine to make compounds.

the answer is c