Question

The stopcock between the two reaction containers is opened, and the reaction proceeds using proper catalysts. Calculate the partial pressure of NO after the reaction is complete. Assume 100% yield for the reaction, that the final container volume is 3.35 L, and that the temperature is constant.

Answer 1

Answer:

0.3 atm.

Explanation:

The first thing to do is to find the number of moles of Ammonia,NH3 and the number of moles of Oxygen,O2.

From the question, we are given the following parameters; The Initial partial pressure of NH3 = 0.5 atm. The Initial volume and final volume of NH3 is 2L and 3L respectively. Also, The Initial volume and final volume of O2 is 1L and 3L respectively, the Initial partial pressure of O2 is 1.5 atm.

The number of moles,n of NH3= 0.0446 mole and the number of moles of oxygen= 0.067 moles.

The next thing to do is to do is to find the limiting reagent which is ammonia,NH3.

We can now use the limiting reagent to find the partial pressure of NO.

So, PV = nRT.

P × 3= 0.0446 × 0.0806 × 273.

P= 0.0446 × 0.08206 × 273/ 3.

P= 0.999146148/3.

Partial pressure of NO= 0.3 atm.

OR

We can just find the partial pressure directly from NH3 as below;

0.5 × 2/ 3 = 0.33 atm.

From the balanced equation, 4 moles of NH3 produces 4 moles of NO.

So, 0.33 atm for NH3 = 0.33 atm for NO.